On the maximal deviation of kernel regression estimators with NMAR response variables

Abstract

This article focuses on the problem of kernel regression estimation in the presence of nonignorable incomplete data with particular focus on the limiting distribution of the maximal deviation of the proposed estimators. From an applied point of view, such a limiting distribution enables one to construct asymptotically correct uniform bands, or perform tests of hypotheses, for a regression curve when the available data set suffers from missing (not necessarily at random) response values. Furthermore, such asymptotic results have always been of theoretical interest in mathematical statistics. We also present some numerical results that further confirm and complement the theoretical developments of this paper.

Appendix: Proofs

To prove our main results, we first state a number of lemmas.

Lemma 1

Let ${\widetilde{\pi }}_{{\widehat{\gamma }}}(x, y)$ be the estimator obtained from ${\widetilde{\pi }}_{\gamma }(x, y)$ upon replacing $\gamma$ by any estimator ${\widehat{\gamma }}$ in (9). Then, under the conditions of Theorem 2, one has

$$\begin{aligned}&\sup _{x\in [0,1]}\, \max _{1\le i \le n} \left| \frac{1}{{\widetilde{\pi }}_{{\widehat{\gamma }}}(X_i, Y_i)} - \frac{1}{{\widetilde{\pi }}_{\gamma }(X_i, Y_i)}\right| \cdot \mathbf{I}\big \{x-Ah_n \le X_i \le x+A h_n\big \}\nonumber \\&\quad = o_p\bigg (\frac{1}{\sqrt{n h_n \log n}}\bigg ) \end{aligned}$$

(25)

$$\begin{aligned}&\sup _{x\in [0,1]}\, \max _{1\le i \le n} \left| \frac{1}{{\widetilde{\pi }}_{\gamma }(X_i, Y_i)} - \frac{1}{\pi _{\gamma }(X_i, Y_i)}\right| \cdot \mathbf{I}\big \{x-Ah_n \le X_i \le x+A h_n\big \}\nonumber \\&\quad = {{\mathcal {O}}}_p\bigg (\sqrt{\frac{\log n}{n \lambda _n}}\bigg ) \end{aligned}$$

(26)

Lemma 2

Let ${\widetilde{m}}_{\pi ,n}(x)$ and ${\widehat{m}}_{n}(x)$ be as in (8) and (11), respectively. Then,

$$\begin{aligned} \sup _{x\in [0, 1]} \Big |{\widehat{m}}_{n}(x) - {\widetilde{m}}_{\pi ,n}(x)\Big |= & {} o_p\bigg (\frac{1}{\sqrt{n h_n \log n}}\bigg ) + {{\mathcal {O}}}_p \bigg (\sqrt{\frac{\log n}{n \lambda _n}} \bigg ). \end{aligned}$$

(27)

To state our next lemma, we first need to define the following auxiliary quantities, which may be viewed as particular estimates of $\nu ^2(x)$ defined in (14)

$$\begin{aligned} {\widetilde{\nu }}^2_{\pi }(x)= & {} \sum _{i=1}^n \, \bigg \{\left[ \frac{\Delta _i Y_i}{\pi _{\gamma }(X_i, Y_i)}+\varepsilon _i\right] ^2\, {{\mathcal {K}}}\left( \frac{x-X_i}{h_n}\right) \bigg \} \Big / \sum _{i=1}^n {{\mathcal {K}}}\left( \frac{x-X_i}{h_n}\right) \nonumber \\&- \big [{\widetilde{m}}_{\pi ,n}(x)\big ]^2 \end{aligned}$$

(28)

$$\begin{aligned} {\widetilde{\nu }}^2_{{\widetilde{\pi }}}(x)= & {} \sum _{i=1}^n \, \bigg \{\left[ \frac{\Delta _i Y_i}{{\widetilde{\pi }}_{\gamma }(X_i, Y_i)}+\varepsilon _i\right] ^2\, {{\mathcal {K}}}\left( \frac{x-X_i}{h_n}\right) \bigg \} \Big / \sum _{i=1}^n {{\mathcal {K}}}\left( \frac{x-X_i}{h_n}\right) \nonumber \\&- \big [{\widetilde{m}}_{{\widetilde{\pi }},n}(x)\big ]^2, \end{aligned}$$

(29)

where ${\widetilde{\pi }}_{\gamma }(x, y)$ is as in (9) and

$$\begin{aligned} {\widetilde{m}}_{{\widetilde{\pi }},n}(x)= \sum _{i=1}^n \, \left\{ \left[ \frac{\Delta _i Y_i}{{\widetilde{\pi }}_{\gamma }(X_i,Y_i)}+\varepsilon _i\right] \, {\mathcal {K}}\left( \frac{x-X_i}{h_n}\right) \right\} \Big /\,\sum _{i=1}^n {\mathcal {K}}\left( \frac{x-X_i}{h_n}\right) . \end{aligned}$$

(30)

Lemma 3

Let ${\widehat{\nu }}^2_{{\widetilde{\pi }}}(x)$, $\nu ^2(x)$, ${\widetilde{\nu }}^2_{{\widetilde{\pi }}}(x)$, and ${\widetilde{\nu }}^2_{\pi }(x)$ be as in (13), (14), (29), and (28), respectively. Then

$$\begin{aligned} \sup _{x\in [0, 1]} \Big |{\widehat{\nu }}^2_{{\widetilde{\pi }}}(x) - {\widetilde{\nu }}^2_{{\widetilde{\pi }}}(x) \Big |= & {} o_p\big (1/\sqrt{n h_n \log n}\,\big ), \end{aligned}$$

(31)

$$\begin{aligned} \sup _{x\in [0, 1]} \Big |{\widetilde{\nu }}^2_{{\widetilde{\pi }}}(x) - {\widetilde{\nu }}^2_{\pi }(x) \Big |= & {} {{\mathcal {O}}}_p \big (\sqrt{(n \lambda _n)^{-1}\log n} \big ), \end{aligned}$$

(32)

$$\begin{aligned} \sup _{x\in [0, 1]} \Big |{\widetilde{\nu }}^2_{\pi }(x) - \nu ^2(x) \Big |= & {} {{\mathcal {O}}}_p \big (\sqrt{(n h_n)^{-1}\log n}\bigg ). \end{aligned}$$

(33)

Proof of Theorem 2

To prove Theorem 2, we first consider the following simple decomposition

$$\begin{aligned} \sup _{x\in [0,1]}\, \sqrt{\frac{f_n(x)}{{\widehat{\nu }}^2_{{\widetilde{\pi }}}(x)}}~ \bigg | {\widehat{m}}_n(x)-m(x)\bigg |= & {} \sup _{x\in [0,1]}\, \sqrt{\frac{f_n(x)}{{\widetilde{\nu }}^2_{\pi }(x)}}~ \bigg | {\widetilde{m}}_{\pi ,n}(x)-m(x)\bigg | \,+\, {{\mathcal {R}}}_n \nonumber \\ \end{aligned}$$

(34)

where the remainder term, ${{\mathcal {R}}}_n$, is given by

$$\begin{aligned} {{\mathcal {R}}}_n&= \sup _{x\in [0,1]}\, \sqrt{\frac{f_n(x)}{ {\widehat{\nu }}^2_{{\widetilde{\pi }}}(x)}}~ \bigg | {\widehat{m}}_n(x)-m(x)\bigg | - \sup _{x\in [0,1]}\, \sqrt{\frac{f_n(x)}{ {\widetilde{\nu }}^2_{\pi }(x)}}~ \bigg | {\widetilde{m}}_{\pi ,n}(x)-m(x)\bigg | \nonumber \\&\le \sup _{x\in [0,1]}\, \sqrt{\frac{f_n(x)}{ {\widehat{\nu }}^2_{{\widetilde{\pi }}}(x)}}~ \bigg | {\widehat{m}}_n(x)-{\widetilde{m}}_{\pi ,n}(x)\bigg |\nonumber \\&\quad + \sup _{x\in [0,1]}\, \sqrt{\frac{{\widetilde{\nu }}^2_{\pi }(x)}{ {\widehat{\nu }}^2_{{\widetilde{\pi }}}(x)}}\, \sqrt{\frac{f_n(x)}{{\widetilde{\nu }}^2_{\pi }(x)}}~ \bigg | {\widetilde{m}}_{\pi ,n}(x) - m(x)\bigg | \nonumber \\&\quad -\sup _{x\in [0,1]}\, \sqrt{\frac{f_n(x)}{{\widetilde{\nu }}^2_{\pi }(x)}}~ \bigg | {\widetilde{m}}_{\pi ,n}(x) - m(x)\bigg | \nonumber \\&\le \sup _{x\in [0,1]}\, \sqrt{\frac{f_n(x)}{ {\widehat{\nu }}^2_{{\widetilde{\pi }}}(x)}}~ \bigg | {\widehat{m}}_n(x)-{\widetilde{m}}_{\pi ,n}(x)\bigg | \nonumber \\&\quad + \left[ \sup _{x\in [0,1]}\, \sqrt{\frac{{\widetilde{\nu }}^2_{\pi }(x)}{ {\widehat{\nu }}^2_{{\widetilde{\pi }}}(x)}}-1\right] \cdot \sup _{x\in [0,1]}\, \sqrt{\frac{f_n(x)}{{\widetilde{\nu }}^2_{\pi }(x)}}~ \bigg | {\widetilde{m}}_{\pi ,n}(x) - m(x)\bigg | \nonumber \\&=: {{\mathcal {R}}}_n(i) + {{\mathcal {R}}}_n(ii) \end{aligned}$$

(35)

To deal with the first term on the r.h.s of (34), first observe that ${\widetilde{m}}_{\pi ,n}(x)$ and ${\widetilde{\nu }}_{\pi ,n}^2(x)$ that appear in this supremum term are, respectively, the kernel regression estimator of $E(Y^*|X=x)$ and the kernel estimator of the conditional variance of $Y^*$ based on the iid “data” $(X_i, Y^*_i),$ $i=1,\ldots ,n$, where $Y^*=\Delta Y\big /\pi _{\gamma }(X,Y) + \varepsilon$; see (12). Furthermore, when assumptions (A), (F), and (G) hold, we have $P\{B_L\le Y^* \le B^U\}=1$ for finite constants $B_L$ and $B^U$. In fact, one can take $B_L=\pi ^{-1}_{\mathrm{min}}\min (0,B_1)+a_0$  and $B^U=\pi ^{-1}_{\mathrm{min}}B_2+b_0$, where $B_1$ and $B_2$ are the constants in Assumption (A), the term $\pi _{\mathrm{min}}$ is as in assumption (F), and $a_0$ and $b_0$ are given in Assumption (G). Therefore, when Assumption (A) holds for the distribution of (X, Y) then, in view of assumptions (F) and (G), it also holds for the distribution of $(X,Y^*)$ with $B_1$ and $B_2$ replaced by $B_L$ and $B^U$. Additionally, it is not hard to show that, in view of Assumption (F), if $\nu _0^2(x) := E[(Y-m(X))^2|X=x]$ satisfies Assumption (C) then so does $\nu ^2(x)$. Hence, in view of Theorem 1, and under assumptions (A), (B), (C), (D$'$), (E$'$), (F), and (G), the first term on the r.h.s of (34) satisfies

$$\begin{aligned}&P\left\{ \sqrt{2 \delta \log n}\left( \sqrt{\frac{n h_n}{c_K}}\, \sup _{x\in [0,1]} \sqrt{\frac{f_n(x)}{{\widetilde{\nu }}^2_{\pi }(x)}}\, \Big |{\widetilde{m}}_{\pi ,n}(x)-m(x)\Big |- \varphi (n)\right) \le u\right\} \nonumber \\&\quad \rightarrow \exp \left( -2e^{-u}\right) \end{aligned}$$

(36)

where $c_K=\int K^2(t)\,dt$ and $\varphi (n)$ is as in (15). Now to finish the proof of Theorem 2, we have to show that $\sqrt{n h_n \log n}\,{{\mathcal {R}}}_n\rightarrow ^p 0$, as $n\rightarrow \infty$. However, by (35), it is sufficient to show that $\sqrt{n h_n \log n}\,\big |{{\mathcal {R}}}_n(i)\big |\rightarrow ^p 0$ and $\sqrt{n h_n \log n}\,\big |{{\mathcal {R}}}_n(ii)\big |\rightarrow ^p 0$. To this end, first note that (36) yields

$$\begin{aligned} \sup _{x\in [0,1]} \sqrt{\frac{f_n(x)}{{\widetilde{\nu }}^2_{\pi }(x)}}\, \Big | {\widetilde{m}}_{\pi ,n}(x) - m(x)\Big |\,=\,{{\mathcal {O}}}_p \bigg (\sqrt{\frac{\log n}{n h_n}}\bigg ). \end{aligned}$$

(37)

We also note that

$$\begin{aligned} \left| \sup _{x\in [0,1]}\, \sqrt{\frac{{\widetilde{\nu }}^2_{\pi }(x)}{ {\widehat{\nu }}^2_{{\widetilde{\pi }}}(x)}}-1\right|\le & {} \sup _{x\in [0,1]}\left| \, \sqrt{\frac{{\widetilde{\nu }}^2_{\pi }(x)}{ {\widehat{\nu }}^2_{{\widetilde{\pi }}}(x)}}-1\right| ~\le \, \sup _{x\in [0,1]}\frac{\big |{\widetilde{\nu }}^2_{\pi }(x)- {\widehat{\nu }}^2_{{\widetilde{\pi }}}(x)\big |}{{\widehat{\nu }}^2_{{\widetilde{\pi }}}(x)}. \end{aligned}$$

(38)

However, in view of (32) and (31),

$$\begin{aligned} \sup _{x\in [0,1]}\big |{\widetilde{\nu }}^2_{\pi }(x)- {\widehat{\nu }}^2_{{\widetilde{\pi }}}(x)\big |= & {} o_p\bigg (\frac{1}{\sqrt{n h_n \log n}}\bigg ) + {{\mathcal {O}}}_p \bigg (\sqrt{\frac{\log n}{n \lambda _n}} \bigg ). \end{aligned}$$

(39)

Also, observe that

$$\begin{aligned} \inf _{x\in [0,1]} {\widehat{\nu }}^2_{{\widetilde{\pi }}}(x)\ge & {} -\sup _{x\in [0, 1]} \big |{\widehat{\nu }}^2_{{\widetilde{\pi }}}(x) - {\widetilde{\nu }}^2_{{\widetilde{\pi }}}(x) \big | - \sup _{x\in [0, 1]} \big |{\widetilde{\nu }}^2_{{\widetilde{\pi }}}(x) - {\widetilde{\nu }}^2_{\pi }(x) \big | \nonumber \\&- \sup _{x\in [0, 1]} \Big |{\widetilde{\nu }}^2_{\pi }(x) - \nu ^2(x) \Big | + \inf _{x\in [0,1]}\nu ^2(x) \end{aligned}$$

(40)

$$\begin{aligned} \inf _{x\in [0,1]} {\widehat{\nu }}^2_{{\widetilde{\pi }}}(x)\le & {} \sup _{x\in [0, 1]} \big |{\widehat{\nu }}^2_{{\widetilde{\pi }}}(x) - {\widetilde{\nu }}^2_{{\widetilde{\pi }}}(x) \big | + \sup _{x\in [0, 1]} \big |{\widetilde{\nu }}^2_{{\widetilde{\pi }}}(x) - {\widetilde{\nu }}^2_{\pi }(x) \big | \nonumber \\&+ \sup _{x\in [0, 1]} \Big |{\widetilde{\nu }}^2_{\pi }(x) - \nu ^2(x) \Big | + \inf _{x\in [0,1]}\nu ^2(x). \end{aligned}$$

(41)

Now, taking the limit, as $n\rightarrow \infty$, of both sides of (40) and (41) and taking into account Lemma 3, we arrive at

$$\begin{aligned} 0< \lim _{n\rightarrow \infty } \inf _{x\in [0,1]} {\widehat{\nu }}^2_{{\widetilde{\pi }}}(x) < \infty . \end{aligned}$$

(42)

This together with (39), (38), and (37) yields

$$\begin{aligned} |{{\mathcal {R}}}_n(ii)| \,=\, {{\mathcal {O}}}_p \bigg (\sqrt{\frac{\log n}{n h_n}}\bigg ) \left[ o_p\bigg (\frac{1}{\sqrt{n h_n \log n}}\bigg ) + {{\mathcal {O}}}_p \bigg (\sqrt{\frac{\log n}{n \lambda _n}} \bigg ) \right] , \end{aligned}$$

from which we arrive at

$$\begin{aligned} \sqrt{n h_n \log n}\,\big |{{\mathcal {R}}}_n(ii)\big | \,=\, o_p \bigg (\sqrt{\frac{\log n}{n h_n}}\,\bigg ) + {{\mathcal {O}}}_p \left( \frac{(\log n)^{3/2}}{\sqrt{n \lambda _n}}\right) \,=\, o_p(1). \end{aligned}$$

To deal with the term ${{\mathcal {R}}}_n(i)$ in (35), first note that by Lemma 2

$$\begin{aligned}&\sqrt{n h_n \log n}\, \sup _{x\in [0,1]}\, \Big | {\widehat{m}}_n(x)-{\widetilde{m}}_{\pi ,n}(x)\Big | \\&\quad = \sqrt{n h_n \log n}\,\left[ {{\mathcal {O}}}_p \bigg (\sqrt{\frac{\log n}{n \lambda _n}} \bigg ) + o_p\bigg (\frac{1}{\sqrt{n h_n \log n}}\bigg ) \right] \\&\quad = {{\mathcal {O}}}_p\Big (\sqrt{n^{\beta -\delta }\, (\log n)^{3/2}}\Big ) + o_p(1) = o_p(1), \end{aligned}$$

where we have used the fact that $\beta <\delta$. Furthermore, since by (42), $\sup _{x\in [0,1]}\, \big | f_n(x)/ {\widehat{\nu }}^2_{{\widetilde{\pi }}} (x)\big | \le \, \big \{\sup _{x\in [0,1]}\, \big | f_n(x)-f(x)\big |+\sup _{x\in [0,1]}f(x)\big \}\big / \inf _{x\in [0,1]}{\widehat{\nu }}^2_{{\widetilde{\pi }}}(x) \,=\, {{\mathcal {O}}}_p(1),$ one finds

$$\begin{aligned} \sqrt{n h_n \log n\,}\,\big |{{\mathcal {R}}}_n(i)\big | = o_p(1). \end{aligned}$$

This completes the proof of Theorem 2. $\square$

Proof of Theorem 3

The proof is similar to that of Theorem 2, but uses a result of Konakov and Piterbarg (1984, Theorem 1.1) instead of that of Liero (1982). $\square$

Proof of Lemma 1

We start by defining the following quantities

$$\begin{aligned} {\widehat{\phi }}_1(x)= & {} \sum _{j=1}^n \big [1-(\Delta _j+\varepsilon _j)\big ] {\mathcal {K}}\left( \frac{x - X_j}{\lambda _n}\right) \Big /\,\sum _{j=1}^n {\mathcal {K}} \left( \frac{x - X_j}{\lambda _n}\right) \end{aligned}$$

(43)

$$\begin{aligned} {\widehat{\phi }}_2(x)= & {} \sum _{j=1}^n (\Delta _j+\varepsilon _j)\exp \{{\widehat{\gamma }}Y_j\}\, {\mathcal {K}}\left( \frac{x - X_j}{\lambda _n}\right) \Big /\,\sum _{j=1}^n {\mathcal {K}} \left( \frac{x - X_j}{\lambda _n}\right) \end{aligned}$$

(44)

$$\begin{aligned} {\widetilde{\phi }}_2(x)= & {} \sum _{j=1}^n (\Delta _j+\varepsilon _j)\exp \{\gamma Y_j\}\, {\mathcal {K}}\left( \frac{x - X_j}{\lambda _n}\right) \Big / \,\sum _{j=1}^n {\mathcal {K}} \left( \frac{x - X_j}{\lambda _n}\right) . \end{aligned}$$

(45)

$$\begin{aligned} \phi _2(x)= & {} E\big [(\Delta +\varepsilon )\exp \{\gamma Y\}\big |X=x\big ]. \end{aligned}$$

(46)

$$\begin{aligned} \phi _1(x)= & {} E\big [1-(\Delta +\varepsilon )\big |X=x\big ]. \end{aligned}$$

(47)

Then it is straightforward to see

$$\begin{aligned}&\left| \frac{1}{{\widetilde{\pi }}_{{\widehat{\gamma }}}(x, Y_i)} - \frac{1}{{\widetilde{\pi }}_{\gamma }(x, Y_i)}\right| \nonumber \\&\quad =\left| \frac{-\exp \{{\widehat{\gamma }}Y_i\} {\widehat{\phi }}_1(x)}{{\widehat{\phi }}_2(x)} \cdot \frac{{\widehat{\phi }}_2(x)-{\widetilde{\phi }}_2(x)}{{\widetilde{\phi }}_2(x)} + \frac{\big [\exp \{{\widehat{\gamma }}Y_i\} - \exp \{\gamma Y_i\}\big ]\, {\widehat{\phi }}_1(x)}{{\widetilde{\phi }}_2(x)}\right| \nonumber \\&\quad \le \left| \frac{1}{{\widetilde{\phi }}_2(x)}\right| \,\left[ \left| \frac{\exp \{{\widehat{\gamma }}Y_i\} {\widehat{\phi }}_1(x)}{{\widehat{\phi }}_2(x)}\right| \cdot \left| {\widehat{\phi }}_2(x) -{\widetilde{\phi }}_2(x)\right| \right. \nonumber \\&\quad \left. + \left| \big [\exp \{{\widehat{\gamma }}Y_i\} - \exp \{\gamma Y_i\}\big ] {\widehat{\phi }}_1(x) \right| \right] ~~~~~ \end{aligned}$$

(48)

Now, put $c :=\max (|B_1|, |B_2|)$, where $B_1$ and $B_2$ are as in Assumption (A), and observe that a one-term Taylor expansion gives

$$\begin{aligned} \left| {\widehat{\phi }}_2(x)-{\widetilde{\phi }}_2(x)\right|= & {} \left| \frac{\sum _{j=1}^n (\Delta _j+\varepsilon _j) \big [\exp \{{\widehat{\gamma }}Y_j\}-\exp \{\gamma Y_j\}\big ]\, {\mathcal {K}}\left( \frac{x - X_j}{\lambda _n}\right) }{\sum _{j=1}^n {\mathcal {K}} \left( \frac{x - X_j}{\lambda _n}\right) }\right| \nonumber \\\le & {} \left| \frac{c\sum _{j=1}^n (1+|\varepsilon _j|)|{\widehat{\gamma }}-\gamma |\exp \big \{|\gamma ^*-\gamma |c+ \gamma c\big \} \, {\mathcal {K}}\left( \frac{x - X_j}{\lambda _n}\right) }{\sum _{j=1}^n {\mathcal {K}} \left( \frac{x - X_j}{\lambda _n}\right) }\right| ,\nonumber \\[2pt]&(\gamma ^* \text{ is } \text{ a } \text{ point } \text{ on } \text{ the } \text{ interior } \text{ of } \text{ the } \text{ line } \text{ joining } {\widehat{\gamma }} \text{ and } \gamma )\nonumber \\\le & {} c\,\big (1+|a_0|\vee b_0\big ) |{\widehat{\gamma }}-\gamma |\exp \big \{ |\gamma ^*- \gamma |c+\gamma c\big \}\nonumber \\&(\text{ where } a_0 \text{ and } b_0 \text{ are } \text{ as } \text{ in } \text{ Assumption } \text{(G) })\nonumber \\\le & {} c_0 |{\widehat{\gamma }}-\gamma |\exp \big \{c\big [\gamma +|{\widehat{\gamma }} - \gamma |c\big \},\nonumber \\&\quad \text{ where } c_0 = c\,\big (1+|a_0|\vee b_0\big ), \nonumber \\= & {} o_p\bigg (\frac{1}{\sqrt{n h_n \log n}}\bigg ) \cdot {{\mathcal {O}}}_p \left( 1\right) , \end{aligned}$$

(49)

where the bound does not depend on x. Similarly, we note that

$$\begin{aligned}&\bigg |\Big [\exp \{{\widehat{\gamma }}Y_i\} - \exp \{\gamma Y_i\}\Big ]\, {\widehat{\phi }}_1(x)\bigg | \nonumber \\&\quad \le \Big |\exp \{{\widehat{\gamma }}Y_i\}-\exp \{\gamma Y_i\}\Big |\,\frac{\sum _{j=1}^n \big |1-(\Delta _j+\varepsilon _j)\big | \, {\mathcal {K}}\left( \frac{x - X_j}{\lambda _n}\right) }{\sum _{j=1}^n {\mathcal {K}} \left( \frac{x - X_j}{\lambda _n}\right) }\nonumber \\&\quad \le \big (2+|a_0|\vee b_0\big )c\,|{\widehat{\gamma }}-\gamma |\exp \big \{ c\big [\gamma +|{\widehat{\gamma }}- \gamma |\big ]\big \} \nonumber \\&\quad = o_p\bigg (\frac{1}{\sqrt{n h_n \log n}}\bigg ) \cdot {{\mathcal {O}}}_p \left( 1\right) , \end{aligned}$$

(50)

where the bound in (50) does not depend on the particular x or $Y_i$. Now, observe that

$$\begin{aligned}&\sup _{x\in [0,1]}\, \max _{1\le i \le n} \left| \frac{1}{{\widetilde{\pi }}_{{\widehat{\gamma }}}(X_i, Y_i)} - \frac{1}{{\widetilde{\pi }}_{\gamma }(X_i, Y_i)}\right| \cdot \mathbf{I}\big \{x-Ah_n \le X_i \le x+A h_n\big \} \nonumber \\&\quad \le \max _{1\le i \le n} \sup _{-h_n\le \, x\, \le 1+A h_n} \Bigg \{\left| \frac{1}{{\widetilde{\phi }}_2(x)}\right| \cdot \left[ \Bigg | \frac{\exp \{{\widehat{\gamma }}Y_i\}\,{\widehat{\phi }}_1(x)}{{\widehat{\phi }}_2(x)}\right| \cdot \left| {\widehat{\phi }}_2(x)-{\widetilde{\phi }}_2(x)\right| \nonumber \\&\qquad + \Big |\big [\exp \{{\widehat{\gamma }}Y_i\} - \exp \{\gamma Y_i\}\big ]\, {\widehat{\phi }}_1(x)\Big |\Bigg ]\Bigg \}. \end{aligned}$$

(51)

To deal with the right side of (51), first note that

$$\begin{aligned}&\sup _{-h_n\, \le x\, \le 1+A h_n} \left| \frac{\exp \{{\widehat{\gamma }}Y_i\}\,{\widehat{\phi }}_1(x)}{{\widehat{\phi }}_2(x)}\right| \\&\quad \le \sup _{-h_n\le x \le 1+A h_n} \bigg \{\bigg [ \left| \big [\exp \{{\widehat{\gamma }}Y_i\} - \exp \{\gamma Y_i\}\big ]\, {\widehat{\phi }}_1(x) \right| + \left| \big [{\widehat{\phi }}_1(x) - \phi _1(x)\big ] \exp \{\gamma Y_i\}\right| \nonumber \\&\qquad + \big |\phi _1(x) \exp \{\gamma Y_i\}\big |\bigg ] \Big /\, \left| {\widehat{\phi }}_2(x) \right| \bigg \} \nonumber \end{aligned}$$

(52)

Now, since the bound in (50) does not depend on any particular x or $Y_i$, one finds

$$\begin{aligned} \sup _{x\in [0,1]}\, \max _{1\le i \le n} \left| \big [\exp \{{\widehat{\gamma }}Y_i\} - \exp \{\gamma Y_i\}\big ]\, {\widehat{\phi }}_1(x) \right|= & {} o_p\bigg (\frac{1}{\sqrt{n h_n \log n}}\bigg ). \end{aligned}$$

(53)

Next, let n be large enough so that $A h_n <\epsilon$, where $\epsilon$ is as in assumption (B), and observe that by the results of Mack and Silverman (1982; Theorem B), one has

$$\begin{aligned} \sup _{x\in [0,1]}\, \max _{1\le i \le n} \left| \big [{\widehat{\phi }}_1(x) -\phi _1(x)\big ] \exp \{\gamma Y_i\}\right|\le & {} {{\mathcal {O}}}_p \bigg (\sqrt{\frac{\log n}{n \lambda _n}} \bigg )\times \exp \{\gamma c\} \nonumber \\= & {} {{\mathcal {O}}}_p \bigg (\sqrt{\frac{\log n}{n \lambda _n}} \bigg ), \end{aligned}$$

(54)

where $c :=\max (|B_1|, |B_2|)$ as before, and $B_1$ and $B_2$ are as in assumption (A). Furthermore,

$$\begin{aligned} \sup _{x\in [0,1]}\, \max _{1\le i \le n} \left| \phi _1(x) \exp \{\gamma Y_i\}\right| ~\le ~ (1-\pi _{\mathrm{min}})\exp \{\gamma c\} \,=~ {{\mathcal {O}}}(1). \end{aligned}$$

(55)

We also need to deal with the infimum of the term $\big |{\widehat{\phi }}_2(x) \big |$ that appears in the denominator of (52). To this end, we first note that $\big |{\widehat{\phi }}_2(x) \big |$ can be upper- and lower-bounded as follows

$$\begin{aligned}&\left| \phi _2(x)\right| - \left| {\widetilde{\phi }}_2(x) - \phi _2(x)\right| -\left| {\widehat{\phi }}_2(x)-{\widetilde{\phi }}_2(x)\right| \\&\quad \le \big |{\widehat{\phi }}_2(x)\big | \le \left| {\widehat{\phi }}_2(x)-{\widetilde{\phi }}_2(x)\right| + \left| {\widetilde{\phi }}_2(x) - \phi _2(x)\right| + \left| \phi _2(x)\right| \end{aligned}$$

Taking the infimum over $x\in [-h_n,\, 1+A h_n]$, we find  $\inf _x\left| \phi _2(x)\right| - \sup _x\big |{\widetilde{\phi }}_2(x) - \phi _2(x)\big | -\sup _x\big |{\widehat{\phi }}_2(x) -{\widetilde{\phi }}_2(x)\big |\le ~\inf _x \big |{\widehat{\phi }}_2(x)\big | \,\le \, \sup _x\big |{\widehat{\phi }}_2(x)-{\widetilde{\phi }}_2(x)\big |+ \sup _x\big |{\widetilde{\phi }}_2(x) - \phi _2(x)\big |+ \sup _x\big |\phi _2(x)\big |.$ Therefore, taking the limit as $n\rightarrow \infty$, one finds

$$\begin{aligned} 0 < \varphi _0 \le \lim _{n\rightarrow \infty } \inf _{-h_n\, \le x\, \le 1+A h_n} \big |{\widehat{\phi }}_2(x)\big |~ \le ~ \exp \{\gamma c\}, \end{aligned}$$

(56)

for a positive constant $\varphi _0$ not depending on n. Here, (56) follows from (49) in conjunction with Theorem B of Mack and Silverman (1982). Furthermore, similar (and in fact easier) arguments can also be used to show that

$$\begin{aligned} 0 < \varphi _0 \le \lim _{n\rightarrow \infty } \inf _{-h_n\, \le x\, \le 1+A h_n} \big |{\widetilde{\phi }}_2(x)\big | ~\le ~ \exp \{\gamma c\}. \end{aligned}$$

(57)

Now (25) follows from (57), (56), (55), (54), (53), (51), and (48). The proof of (26) is very similar to (and, in fact, easier than) that of (25) and therefore will not be given. $\square$

Proof of Lemma 2

Let ${\widetilde{m}}_{{\widetilde{\pi }},n}(x)$ be as in (30), and note that

$$\begin{aligned}&\Big |{\widetilde{m}}_{{\widetilde{\pi }},n}(x) - {\widetilde{m}}_{\pi ,n}(x)\Big |\\&\quad = \left| \frac{ \sum _{j=1}^n (\Delta _j+\varepsilon _j)Y_j \left[ \frac{1}{{\widetilde{\pi }}_{\gamma }(X_i, Y_i)} - \frac{1}{\pi _{\gamma }(X_i, Y_i)}\right] {\mathcal {K}}\left( \frac{x - X_j}{h_n}\right) }{\sum _{j=1}^n {\mathcal {K}} \left( \frac{x - X_j}{h_n}\right) }\right| \\&\quad \le \max _{1\le i \le n}\left\{ \left| \frac{1}{{\widetilde{\pi }}_{\gamma }(X_i, Y_i)} - \frac{1}{\pi _{\gamma }(X_i, Y_i)}\right| \, \mathbf{I}\big \{x-Ah_n \le X_i \le x+A h_n\big \}\right\} \\&\qquad \times \left[ \sum _{j=1}^n \Big |(\Delta _j+\varepsilon _j)Y_j\Big |{\mathcal {K}} \left( \frac{x - X_j}{h_n}\right) \Big /\,\sum _{j=1}^n {\mathcal {K}}\left( \frac{x- X_j}{h_n}\right) \right] \\&\quad \le c_2 \max _{1\le i \le n}\left\{ \left| \frac{1}{{\widetilde{\pi }}_{\gamma }(X_i, Y_i)} - \frac{1}{\pi _{\gamma }(X_i, Y_i)}\right| \mathbf{I}\big \{x-Ah_n \le X_i \le x+A h_n\big \}\right\} ,~ \end{aligned}$$

where $c_2$ is a positive constant not depending on n. Therefore, in view of (26),

$$\begin{aligned} \sup _{x\in [0,1]}\,\Big |{\widetilde{m}}_{{\widetilde{\pi }},n}(x) - {\widetilde{m}}_{\pi ,n}(x)\Big | ~=~ {{\mathcal {O}}}_p\bigg (\sqrt{\frac{\log n}{n \lambda _n}}\bigg ). \end{aligned}$$

(58)

Similarly, one has

$$\begin{aligned}&\Big |{\widehat{m}}_{n}(x) - {\widetilde{m}}_{{\widetilde{\pi }},n}(x)\Big | \\&\quad = \left| \frac{ \sum _{j=1}^n (\Delta _j+\varepsilon _j)Y_j \left[ \frac{1}{{\widetilde{\pi }}_{{\widehat{\gamma }}}(X_j, Y_j)} - \frac{1}{{\widetilde{\pi }}_{\gamma }(X_j, Y_j)}\right] {\mathcal {K}}\left( \frac{x - X_j}{h_n}\right) }{\sum _{j=1}^n {\mathcal {K}} \left( \frac{x - X_j}{h_n}\right) }\right| \\&\quad \le c_2 \max _{1\le i \le n}\left\{ \left| \frac{1}{{\widetilde{\pi }}_{{\widehat{\gamma }}}(X_i, Y_i)} - \frac{1}{{\widetilde{\pi }}_{\gamma }(X_i, Y_i)}\right| \mathbf{I}\big \{x-Ah_n \le X_i \le x+A h_n\big \}\right\} ,~ \end{aligned}$$

which, together with (25), yields

$$\begin{aligned} \sup _{x\in [0,1]}\, \Big |{\widehat{m}}_{n}(x) - {\widetilde{m}}_{{\widetilde{\pi }},n}(x)\Big | = o_p\bigg (\frac{1}{\sqrt{n h_n \log n}}\bigg ). \end{aligned}$$

(59)

The proof of Lemma 2 now follows from (58) and (59) and the fact that $\big |{\widehat{m}}_{n}(x) - {\widetilde{m}}_{\pi ,n}(x)\big | \le \big |{\widehat{m}}_{n}(x) - {\widetilde{m}}_{{\widetilde{\pi }},n}(x)\big | + \big |{\widetilde{m}}_{{\widetilde{\pi }},n}(x) - {\widetilde{m}}_{\pi ,n}(x)\big |.$ $\square$

Proof of Lemma 3

We start with the proof of (31). First observe that

$$\begin{aligned}&\Big |{\widehat{\nu }}^2_{{\widetilde{\pi }}}(x) - {\widetilde{\nu }}^2_{{\widetilde{\pi }}}(x) \Big |\nonumber \\&\quad \le \left| \left[ \sum _{i=1}^n \Delta _i Y_i^2 \left[ \frac{1}{[{\widetilde{\pi }}_{{\widehat{\gamma }}}(X_i, Y_i)]^2} - \frac{1}{[{\widetilde{\pi }}_{\gamma }(X_i, Y_i)]^2}\right] \right. \right. \nonumber \\&\qquad \left. \left. \times {{\mathcal {K}}}\left( \frac{x-X_i}{h_n}\right) \right] \Big /\sum _{i=1}^n{{\mathcal {K}}}\left( \frac{x-X_i}{h_n}\right) \right| \nonumber \\&\qquad + 2 \left| \left[ \sum _{i=1}^n\varepsilon _i \, \Delta _i Y_i \left[ \frac{1}{{\widetilde{\pi }}_{{\widehat{\gamma }}}(X_i, Y_i)} - \frac{1}{{\widetilde{\pi }}_{\gamma }(X_i, Y_i)}\right] \right. \right. \nonumber \\&\qquad \left. \left. \times {{\mathcal {K}}}\left( \frac{x-X_i}{h_n}\right) \right] \Big / \sum _{i=1}^n{{\mathcal {K}}}\left( \frac{x-X_i}{h_n}\right) \right| \nonumber \\&\qquad +\Big |\big ({\widetilde{m}}_{{\widetilde{\pi }},n}(x) - {\widehat{m}}_{n}(x)\big )\big ({\widetilde{m}}_{{\widetilde{\pi }},n}(x) + {\widehat{m}}_{n}(x)\big )\Big |\nonumber \\&\quad =: \big |U_{n,1}(x)\big | + \big |U_{n,2}(x)\big | + \big |U_{n,3}(x)\big |. \end{aligned}$$

(60)

However, we have

$$\begin{aligned} \big |U_{n,1}(x)\big |\le & {} r_n(x) \cdot \max _{1\le i \le n} \Bigg [ \Bigg | \frac{1}{{\widetilde{\pi }}_{{\widehat{\gamma }}}(X_i, Y_i)} - \frac{1}{{\widetilde{\pi }}_{\gamma }(X_i, Y_i)}\Bigg |\\&\times \Bigg \{\Bigg | \frac{1}{{\widetilde{\pi }}_{{\widehat{\gamma }}}(X_i, Y_i)} - \frac{1}{{\widetilde{\pi }}_{\gamma }(X_i, Y_i)}\Bigg |\\&+ 2\,\Bigg |\frac{1}{{\widetilde{\pi }}_{\gamma }(X_i, Y_i)} - \frac{1}{\pi _{\gamma }(X_i, Y_i)}\Bigg |\\&+2\,\Bigg |\frac{1}{\pi _{\gamma }(X_i, Y_i)}\Bigg |\Bigg \} \mathbf{I}\big \{x-Ah_n \le X_i \le x+A h_n\big \}\Bigg ], \end{aligned}$$

where $r_n(x) =\sum _{i=1}^n \Delta _i Y^2_i\, {{\mathcal {K}}}((x-X_i)/h_n) / \sum _{i=1}^n {{\mathcal {K}}}((x-X_i)/h_n)\le (|B_1|\vee |B_2|)^2$, where $B_1$ and $B_2$ are as in assumption (A). Therefore, in view of (25) and (26), we obtain

$$\begin{aligned} \sup _{x\in [0,1]}\,\big |U_{n,1}(x)\big |= & {} o_p\bigg (\frac{1}{\sqrt{n h_n \log n}}\bigg ) \bigg \{o_p\bigg (\frac{1}{\sqrt{n h_n \log n}}\bigg ) +{{\mathcal {O}}}_p\bigg (\sqrt{\frac{\log n}{n\lambda _n}}\bigg ) + {{\mathcal {O}}}_p(1)\bigg \} \nonumber \\= & {} o_p\bigg (\frac{1}{\sqrt{n h_n \log n}}\bigg ). \end{aligned}$$

Similarly, we have

$$\begin{aligned} \sup _{x\in [0,1]} \,\big |U_{n,2}(x)\big | =o_p\big (1/\sqrt{n h_n \log n}\big ). \end{aligned}$$

Next, to deal with the term $\big |U_{n,3}(x)\big |$ in (60), we observe that $\big |U_{n,3}(x)\big | \le \big |{\widetilde{m}}_{{\widetilde{\pi }},n}(x) - {\widehat{m}}_{n}(x)\big |\times \big \{ \big |{\widetilde{m}}_{{\widetilde{\pi }},n}(x) - {\widehat{m}}_{n}(x)\big | +2 \big |{\widetilde{m}}_{{\widetilde{\pi }},n}(x) - {\widetilde{m}}_{\pi ,n}(x)\big | +2 \big |{\widetilde{m}}_{\pi ,n}(x) - m(x)\big | +2|m(x)|\big \}$. Consequently, in view of (58) and (59) and the result of Mack and Silverman (1982, Theorem B), we get

$$\begin{aligned}&\sup _{x\in [0,1]}\,\big |U_{n,3}(x)\big | \\&\quad = o_p\bigg (\frac{1}{\sqrt{n h_n \log n}}\bigg ) \Bigg \{o_p\bigg (\frac{1}{\sqrt{n h_n \log n}}\bigg ) + {{\mathcal {O}}}_p\bigg (\sqrt{\frac{\log n}{n\lambda _n}}\bigg ) + {{\mathcal {O}}}_p\bigg (\sqrt{\frac{\log n}{n h_n}}\bigg ) + {{\mathcal {O}}}_p(1)\Bigg \} \nonumber \\&\quad = o_p\bigg (\frac{1}{\sqrt{n h_n \log n}}\bigg ). \end{aligned}$$

Now, (31) follows from the above bounds together with (60). The proof of (32) is similar and goes as follows.

$$\begin{aligned}&\Big |{\widetilde{\nu }}^2_{{\widetilde{\pi }}}(x) - {\widetilde{\nu }}^2_{\pi }(x) \Big | \nonumber \\&\quad \le \left| \left[ \sum _{i=1}^n \Delta _i Y_i^2 \left[ \frac{1}{[{\widetilde{\pi }}_{\gamma }(X_i, Y_i)]^2} - \frac{1}{[\pi _{\gamma }(X_i, Y_i)]^2}\right] \right. \right. \nonumber \\&\qquad \left. \left. \times {{\mathcal {K}}}\left( \frac{x-X_i}{h_n}\right) \right] \Big /\sum _{i=1}^n{{\mathcal {K}}}\left( \frac{x-X_i}{h_n}\right) \right| \nonumber \\&\qquad + 2 \left| \left[ \sum _{i=1}^n\varepsilon _i \, \Delta _i Y_i \left[ \frac{1}{{\widetilde{\pi }}_{\gamma }(X_i, Y_i)} - \frac{1}{\pi _{\gamma }(X_i, Y_i)}\right] \right. \right. \nonumber \\&\qquad \left. \left. \times {{\mathcal {K}}}\left( \frac{x-X_i}{h_n}\right) \right] \Big / \sum _{i=1}^n {{\mathcal {K}}}\left( \frac{x-X_i}{h_n}\right) \right| \nonumber \\ \nonumber \\&\qquad +\Big |\big ({\widetilde{m}}_{{\widetilde{\pi }},n}(x) - {\widetilde{m}}_{\pi ,n}(x)\big )\big ({\widetilde{m}}_{{\widetilde{\pi }},n}(x) + {\widetilde{m}}_{\pi ,n}(x)\big )\Big |\nonumber \\&\quad =: \big |T_{n,1}(x)\big | + \big |T_{n,2}(x)\big | + \big |T_{n,3}(x)\big |. \end{aligned}$$

(61)

But

$$\begin{aligned} \big |T_{n,1}(x)\big |\le & {} c_3\max _{1\le i \le n} \Bigg [ \Bigg | \frac{1}{{\widetilde{\pi }}_{\gamma }(X_i, Y_i)} - \frac{1}{\pi _{\gamma }(X_i, Y_i)}\Bigg |\cdot \Bigg \{\Bigg | \frac{1}{{\widetilde{\pi }}_{\gamma }(X_i, Y_i)} - \frac{1}{\pi _{\gamma }(X_i, Y_i)}\Bigg |\\&+ 2\,\bigg |\frac{1}{\pi _{\gamma }(X_i, Y_i)}\bigg |\Bigg \}\Bigg ] \mathbf{I}\big \{x-Ah_n \le X_i \le x+A h_n\big \}, \end{aligned}$$

where $c_3$ is a positive constant not depending on n. Therefore, by (26) and the second part of assumption (F), we have

$$\begin{aligned} \sup _{x\in [0,1]}\,\big |T_{n,1}(x)\big |= & {} {{\mathcal {O}}}_p\bigg (\sqrt{\frac{\log n}{n\lambda _n}}\bigg ) {{\mathcal {O}}}_p\bigg (\sqrt{\frac{\log n}{n\lambda _n}}\bigg ) + \, {{\mathcal {O}}}_p\bigg (\sqrt{\frac{\log n}{n\lambda _n}}\bigg ) {{\mathcal {O}}}(1) \\= & {} {{\mathcal {O}}}_p\bigg (\sqrt{\frac{\log n}{n\lambda _n}} \bigg ). \end{aligned}$$

Similarly, one has $\sup _{x\in [0,1]}\,\big |T_{n,2}(x)\big |= {{\mathcal {O}}}_p \left( \sqrt{\log n/(n\lambda _n)}\right)$. Furthermore, since

$$\begin{aligned} \big |T_{n,2}(x)\big | \le \big |{\widetilde{m}}_{{\widetilde{\pi }},n}(x) - {\widetilde{m}}_{\pi ,n}(x)\big |\Big [\big |{\widetilde{m}}_{{\widetilde{\pi }},n}(x) - {\widetilde{m}}_{\pi ,n}(x)\big | + 2\,\big |{\widetilde{m}}_{\pi ,n}(x)\big |\Big ], \end{aligned}$$

one finds (in view of (58)) $\sup _{x\in [0,1]}\,\big |T_{n,2}(x)\big | = {{\mathcal {O}}}_p \left( \sqrt{\log n/(n\lambda _n)}\right) .$ Now, (32) follows from (61) together with the above bounds. The proof of (33) is straightforward and, in fact, easier than those of (32) and (31), and hence will not be given. $\square$