Nonparametric relative error regression for spatial random variables

Abstract

Let $\displaystyle Z_{\mathbf {i}}=\left( X_{\mathbf {i}},\ Y_{\mathbf {i}}\right) _{\mathbf {i}\in \mathbb {N}^{N}\, N \ge 1}$, be a $\mathbb {R}^d\times \mathbb {R}$-valued measurable strictly stationary spatial process. We consider the problem of estimating the regression function of $Y_{\mathbf {i}}$ given $X_{\mathbf {i}}$. We construct an alternative kernel estimate of the regression function based on the minimization of the mean squared relative error. Under some general mixing assumptions, the almost complete consistency and the asymptotic normality of this estimator are obtained. Its finite-sample performance is compared with a standard kernel regression estimator via a Monte Carlo study and real data example.

Appendix

Proof of Theorem 1 Let

$$\begin{aligned} \widetilde{\theta }(x)=\frac{\widetilde{g_1}(x)}{\widetilde{g_2}(x)} \end{aligned}$$

with

$$\begin{aligned} \widetilde{g_l}(x)=\frac{1}{\widehat{\mathbf {n}}h^d}\sum _{{\mathbf {i}\in \mathcal {I}_\mathbf {n}}}Y^{-l}_\mathbf{i}K(h^{-1}(x-X_\mathbf{i}))\quad \hbox { for }l=1,2. \end{aligned}$$

Next, we use the following decomposition:

$$\begin{aligned} \widetilde{\theta }(x)-\theta (x)=\frac{1}{\widetilde{g_2}(x)}\Big [\widetilde{g_1}(x)-g_1(x)\Big ]+\left[ g_2(x)-\widetilde{g_2}(x)\right] \frac{\theta (x)}{\widetilde{g_2}(x)} \end{aligned}$$

(6)

Thus, Theorem 1 is a consequence of the following intermediate results (cf. Lemmas 1 and 2).

Lemma 1

Under hypotheses (H1), (H2), and (H5), we have, for $l=1,2$, that:

$$\begin{aligned} \left| E\widetilde{g_l}(x)-g_l(x)\right| =O(h^2). \end{aligned}$$

Proof of Lemma 1

By a change of variables, we get, for $l=1,2$,

$$\begin{aligned} E[ \widetilde{g_l}(x)]= & {} \frac{1}{h^d}\int _{{\mathbbm {R}}^d}E[Y^{-l}|X=u)]K\left( \frac{x-u}{h}\right) f(u)\,du\\= & {} \int _{{\mathbbm {R}}^d}g_l (x-hz)K(z)\, dz. \end{aligned}$$

Since both functions f and $r_l$ are of class $\mathcal{C}^2$, we use a Taylor expansion of $g_l(\cdot )$ to write, under (H4)

$$\begin{aligned} \left| E[ \widetilde{g_l}(x)]-g_l(x)\right| \le C h^2. \end{aligned}$$

The last result complete the proof of lemma. $\square$

Lemma 2

Under hypotheses (H3)–(H7), we have, for $l=1,2$, that:

$$\begin{aligned} \sup _{x\in S}|\widetilde{g_l}(x)-E\widetilde{g_l}(x)| = O_{a.co.}\left( \sqrt{\frac{ \log \widehat{\mathbf {n}}}{\widehat{\mathbf {n}} h^d}}\right) . \end{aligned}$$

Proof of Lemma 2

Consider

$$\begin{aligned} \widetilde{g_l}^*(x) =\frac{1}{\widehat{\mathbf {n}}h^d}\sum _{{\mathbf {i}\in \mathcal {I}_\mathbf {n}}}K\Big (h^{-1}(x-X_i)\Big )Y_i^{-l} {\mathbbm {1}}_{|Y_i^{-1}|<\mu _\mathbf{n} }\;\;\text{ with }\; {\mu }_\mathbf{n}=\widehat{\mathbf {n}}^{\gamma /2}. \end{aligned}$$

Therefore, it suffices to prove the following intermediates results

$$\begin{aligned} \sup _{x\in S}\left| E[ \widetilde{g_l}^* (x)]- E[ \widetilde{g_l}(x)]\right| = O\left( \sqrt{\frac{ \log \widehat{\mathbf {n}}}{\widehat{\mathbf {n}}h^d}}\right) , \end{aligned}$$

(7)

$$\begin{aligned} \sup _{x\in S}\left| \widetilde{g_l}^* (x)- \widetilde{g_l} (x)\right| = O_{a. co.}\left( \sqrt{\frac{ \log \widehat{\mathbf {n}}}{\widehat{\mathbf {n}}h^d}}\right) \end{aligned}$$

(8)

and

$$\begin{aligned} \sup _{x\in S} \left| \widetilde{g_l}^* (x)-E[ \widetilde{g_l}^* (x)] \right| = O_{a. co.}\left( \sqrt{\frac{ \log \widehat{\mathbf {n}}}{\widehat{\mathbf {n}}h^d}}\right) . \end{aligned}$$

(9)

• Firstly, for (9), we use the compactness of S to write

$$\begin{aligned} S \subset \bigcup _{j=1}^{d_\mathbf{n}} B(x_k,\tau _\mathbf{n}), \end{aligned}$$

with $d_\mathbf{n}\le \widehat{\mathbf {n}}^{\beta }$ and $\tau _\mathbf{n}\le d_\mathbf{n}^{-1}$ where $\beta =\frac{\delta (d+2)}{2}+\frac{1}{2}+\frac{\gamma }{2}$. So, for all $x\in S$, we pose

$$\begin{aligned} k(x)=\displaystyle \mathrm{arg\min }_{k\in \{1,\ldots d_\mathbf{n}\}}\ \Vert x-x_k\Vert . \end{aligned}$$

Thus, for $l=1,2$,

$$\begin{aligned} \sup _{x\in S}\left| \widetilde{g_l}^*(x) - E\Big [\widetilde{g_l}^*(x)\Big ]\right|\le & {} \underbrace{\sup _{x\in S}\left| \widetilde{g_l}^*(x)-\widetilde{g_l}^*(x_{k(x)})\right| }_{T_1}\\&+\underbrace{\sup _{x\in S} \left| \widetilde{g_l}^*(x_{k(x)})- E\left[ \widetilde{g_l}^*(x_{k(x)})\right] \right| }_{T_2}\\&+ \underbrace{\sup _{x\in S}\left| E\Big [\widetilde{g_l}^*(x_{k(x)})\Big ]- E\Big [\widetilde{g_l}^*(x)\Big ]\right| }_{T_3}. \end{aligned}$$

• Furthermore, for $T_2$, for both $l=1,2$, we have

  $$\begin{aligned} \sup _{x\in S} \left| \widetilde{g_l}^*(x_{k(x)})- E\left[ \widetilde{g_l}^*(x_{k(x)})\right] \right| =\max _{k=1,\ldots d_\mathbf{n}} \left| \widetilde{g_l}^*(x_{k})- E\left[ \widetilde{g_l}^*(x_{k)})\right] \right| \end{aligned}$$

  Thus it suffices to evaluate almost completely

  $$\begin{aligned} \max _{k=1,\ldots d_\mathbf{n}} \left| \widetilde{g_l}^*(x_{k})- E\left[ \widetilde{g_l}^*(x_{k})\right] \right| . \end{aligned}$$

  To do that, we write:

  $$\begin{aligned} \widetilde{g_l}^* (x_k)-E[ \widetilde{g_l}^* (x_k)]= & {} {1\over \widehat{\mathbf {n}}h^d}\sum _{{\mathbf {i}\in \mathcal {I}_\mathbf {n}}}\Delta _{\mathbf {i}} \end{aligned}$$

  where

  $$\begin{aligned} \Delta _\mathbf{i}=Y^{-l}_\mathbf{i}K(h^{-1}(x_k-X_\mathbf{i})){\mathbbm {1}}_{(|Y_\mathbf{i}^{-1}|< \mu _\mathbf{n})}-E\left[ Y^{-l} K(h^{-1}(x_k-X)){\mathbbm {1}}_{(|Y^{-1}|< \mu _\mathbf{n})}\right] . \end{aligned}$$

  Now, similarly to Tran (1990), we use the classical spatial block decomposition for the sum $\displaystyle \sum _{{\mathbf {i}\in \mathcal {I}_\mathbf {n}}}\Delta _{\mathbf {i}}$ as follows

  $$\begin{aligned} U(1,\mathbf {n},\mathbf {j})=\sum _{\begin{array}{c} i_k=2j_kp_{\mathbf {n}}+1\\ k=1,...,N \end{array}}^{2j_kp_{\mathbf {n}}+p_{\mathbf {n}}}\Delta _{\mathbf {i}}, \end{aligned}$$
  $$\begin{aligned} U(2,\mathbf {n},\mathbf {j})=\displaystyle \sum _{\begin{array}{c} i_k=2j_kp_{\mathbf {n}}+1\\ k=1,...,N-1 \end{array}}^{2j_kp_{\mathbf {n}}+p_{\mathbf {n}}} \quad \displaystyle \sum _{i_N=2j_Np_{\mathbf {n}}+p_{\mathbf {n}}+1}^{(j_N+1)p_{\mathbf {n}}}\Delta _{\mathbf {i}}, \end{aligned}$$
  $$\begin{aligned} U(3,\mathbf {n},\mathbf {j})=\displaystyle \sum _{\begin{array}{c} i_k=2j_kp_{\mathbf {n}}+1\\ k=1,...,N-2 \end{array}}^{2j_kp_{\mathbf {n}}+p_{\mathbf {n}}} \quad \displaystyle \sum _{i_{N-1}=2j_{N-1}p_{\mathbf {n}}+p_{\mathbf {n}}+1}^{2(j_{N-1}+1)p_{\mathbf {n}}} \quad \displaystyle \sum _{i_N=2j_Np_{\mathbf {n}}+1}^{2j_Np_{\mathbf {n}}+p_{\mathbf {n}}}\Delta _{\mathbf {i}}, \end{aligned}$$
  $$\begin{aligned} U(4,\mathbf {n},\mathbf {j})=\displaystyle \sum _{\begin{array}{c} i_k=2j_kp_{\mathbf {n}}+1\\ k=1,...,N-2 \end{array}}^{2j_kp_{\mathbf {n}}} \quad \displaystyle \sum _{i_{N-1}=2j_{N-1}p_{\mathbf {n}}+p_{\mathbf {n}}+1}^{2(j_{N-1}+1)p_{\mathbf {n}}}\quad \displaystyle \sum _{i_N=2j_Np_{\mathbf {n}}+p_{\mathbf {n}}+1}^{2(j_N+1)p_{\mathbf {n}}}\Delta _{\mathbf {i}}, \end{aligned}$$

  and so on. Finally

  $$\begin{aligned} U(2^{N-1},\mathbf {n},\mathbf {j})=\displaystyle \sum _{\begin{array}{c} i_k=2j_kp_{\mathbf {n}}+p_{\mathbf {n}}+1\\ k=1,...,N-1 \end{array}}^{2(j_k+1)p_{\mathbf {n}} }\quad \displaystyle \sum _{i_N=2j_Np_{\mathbf {n}}+1}^{2j_Np_{\mathbf {n}}+p_{\mathbf {n}}}\Delta _{\mathbf {i}}, \end{aligned}$$
  $$\begin{aligned} U(2^N,\mathbf {n},\mathbf {j})=\sum _{\begin{array}{c} i_k=2j_kp_{\mathbf {n}}+p_{\mathbf {n}}+1\\ k=1,...,N \end{array}}^{2(j_k+1)p_{\mathbf {n}}} \Delta _{\mathbf {i}} \end{aligned}$$

  (10)

  with $p_{\mathbf {n}}$ is a real sequence will be specified later. Now, we put for all $i = 1, \ldots ,2^N$,

  $$\begin{aligned} T(\mathbf {n},i)=\sum _{\mathbf {j}\in \mathcal {J}}U(i,\mathbf {n},\mathbf {j}). \end{aligned}$$

  (11)

  with $\mathcal {J}=\{0,...,r_1-1\}\times ...\times \{0,...,r_N-1\}$ and $r_l =2 n_lp_{\mathbf {n}}^{-1}$; $l=1,\ldots , N$. Then,

  $$\begin{aligned} \left| \widetilde{g_l}^* (x_k)-E[ \widetilde{g_l}^* (x_k)]\right| =\frac{1}{\widehat{\mathbf {n}}h^d}\displaystyle \sum _{i=1}^{2^N}T(\mathbf {n},i). \end{aligned}$$

  Thus, all it remains to compute

  $$\begin{aligned} \displaystyle {\mathbbm {P}}\left( T(\mathbf {n},i)\ge \eta \widehat{\mathbf {n}}h^d\right) ,\,\qquad \text{ for } \text{ all } i=1,\ldots , 2^N . \end{aligned}$$

  (12)

  Without loss of generality, we will only consider the case $i=1$. For this, we enumerate the $M=\prod _{k=1}^N r_k=2^{-N}\widehat{\mathbf {n}}p_{\mathbf {n}}^{-N}\le \widehat{\mathbf {n}}p_{\mathbf {n}}^{-N}$ random variables $U(1,\mathbf {n},\mathbf {j});\, \mathbf {j}\in \mathcal {J}$ in the arbitrary way $Z_1,\ldots Z_M$. The rest of the proof is very similar to Biau and Cadre (2004) which is based on Lemma 4.5 of Carbon et al. (1997). According this Lemma we can find independent random variables $Z_1^*,\ldots Z_M^*$ has the same low as $Z_{j=1,\ldots M}$ and such that

  $$\begin{aligned} \sum _{j=1}^r E|Z_{\mathbf {j}}-Z_{\mathbf {j}}^*|\le 2C\mu _\mathbf{n} Mp_{\mathbf {n}}^Ns(M-1)p_{\mathbf {n}}^N,p_{\mathbf {n}}^N)\varphi (p_{\mathbf {n}}). \end{aligned}$$

  It follows that

  $$\begin{aligned} \displaystyle {\mathbbm {P}}\left( T(\mathbf {n},1)\ge \eta \widehat{\mathbf {n}}h^d \right) \le {\mathbbm {P}}\left( \left| \sum _{j=1}^MZ_{\mathbf {j}}^*\right| \ge \frac{\eta \widehat{\mathbf {n}}h^d}{2}\right) +{\mathbbm {P}}\left( \sum _{j=1}^M|Z_{\mathbf {j}}-Z_{\mathbf {j}}^*|\ge \frac{\eta \widehat{\mathbf {n}}h^d}{2}\right) . \end{aligned}$$

  Thus, from the Bernstein and Markov inequalities we deduce that

  $$\begin{aligned} \displaystyle B_1:={\mathbbm {P}}\left( \left| \sum _{j=1}^MZ_{\mathbf {j}}^*\right| \ge \frac{M\eta \widehat{\mathbf {n}}h^d}{2M}\right) \le 2\exp \left( -\frac{(\eta \widehat{\mathbf {n}}h^d)^2}{ MVar\left( Z_1^*\right) +Cp_{\mathbf {n}}^N\eta \widehat{\mathbf {n}}h^d}\right) \end{aligned}$$

  and

  $$\begin{aligned} \displaystyle B_2:= & {} {\mathbbm {P}}\left( \sum _{j=1}^M|Z_{\mathbf {j}}-Z_{\mathbf {j}}^*|\ge \frac{\eta \widehat{\mathbf {n}}h^d}{2}\right) \le \frac{2}{\eta \widehat{\mathbf {n}}h^d}\sum _{j=1}^M E|Z_{\mathbf {j}}-Z_{\mathbf {j}}^*|. \end{aligned}$$

  By using Lemma 4.5 of Carbon et al. (2007), the fact that $\displaystyle \widehat{\mathbf {n}}=2^NMp_{\mathbf {n}}^N$ and $s((M-1)p_{\mathbf {n}}^N,p_{\mathbf {n}}^N)\le p_{\mathbf {n}}^N$ we get for $\eta =\displaystyle \eta _0\sqrt{\frac{\log \widehat{\mathbf {n}}}{\widehat{\mathbf {n}}\, h^d}}$

  $$\begin{aligned} B_2\le \mu _\mathbf{n} \widehat{\mathbf {n}} p_{\mathbf {n}}^N\left( \log \widehat{\mathbf {n}}\right) ^{-1/2}\left( \widehat{\mathbf {n}}h^d\right) ^{-1/2}\varphi (p_{\mathbf {n}}). \end{aligned}$$

  Since $p_{\mathbf {n}}= C\left( \frac{\widehat{\mathbf {n}}h^d}{\log \widehat{\mathbf {n}} \mu _\mathbf{n}^2}\right) ^{1/2N}$, then

  $$\begin{aligned} B_2 \le \widehat{\mathbf {n}}\, \left( \log \widehat{\mathbf {n}}\right) ^{-1} \,\varphi (p_{\mathbf {n}}). \end{aligned}$$

  Concerning $B_1$ term, by a standard arguments we obtain

  $$\begin{aligned} Var\left[ Z_1^*\right] =O\left( p_{\mathbf {n}}^Nh^d\right) . \end{aligned}$$

  Using this last result, together with the definitions of $p_{\mathbf {n}}$, M and $\eta$, we get

  $$\begin{aligned} B_1 \le \exp \left( -C(\eta _0)\log \widehat{\mathbf {n}} \right) \end{aligned}$$

  Consequently, from (H6), we have

  $$\begin{aligned} \exists \eta _0 \quad \text{ such } \text{ that } \quad d_\mathbf{n}\sum _{\mathbf {n}}(B_1+B_2)<\infty . \end{aligned}$$

  which complete the first result of this lemma.

• Now, we evaluate terms $T_1$ and $T_3$: To do that, we use the Lipschitz’s condition of the kernel K in (H4) allows to write directly,

  $$\begin{aligned} \left| \widetilde{g_l}^*(x)- \widetilde{g_l}^*(x_{k(x)})\right|= & {} \frac{1}{\widehat{\mathbf {n}}h^d}\left| \sum _{{\mathbf {i}\in \mathcal {I}_\mathbf {n}}}Y_\mathbf{i}^{-l}K_\mathbf{i}(x)- \sum _{{\mathbf {i}\in \mathcal {I}_\mathbf {n}}}Y_\mathbf{i}^{-l}K_\mathbf{i}(x_{k(x)})\right| \\\le & {} \frac{C}{\widehat{\mathbf {n}}h^{d+1}}\Vert x- x_{k(x)}\Vert \sum _{{\mathbf {i}\in \mathcal {I}_\mathbf {n}}}Y_\mathbf{i}^{-l}\\\le & {} \frac{C\tau _\mathbf{n} }{\mu _\mathbf{n}^{l}\widehat{\mathbf {n}}h^{d+1}}\le \frac{C\tau _\mathbf{n} }{\mu _\mathbf{n}\widehat{\mathbf {n}}h^{d+1}}. \end{aligned}$$

By the definition of $\tau _\mathbf{n}$ we obtain

$$\begin{aligned} \sup _{x\in S}\left| \widetilde{g_l}^*(x)- \widetilde{g_l}^*(x_{k(x)})\right| =O\left( \sqrt{\frac{ \log \widehat{\mathbf {n}}}{\widehat{\mathbf {n}}h^d}}\right) \end{aligned}$$

(13)

and

$$\begin{aligned} \sup _{x\in S}\left| E\Big [ \widetilde{g_l}^*(x)\Big ]- E\Big [ \widetilde{g_l}^*(x_{k(x)})\Big ]\right| =O\left( \sqrt{\frac{ \log \widehat{\mathbf {n}}}{\widehat{\mathbf {n}}h^d}}\right) \end{aligned}$$

(14)

• Secondly, we proof (7). Indeed, we have

  $$\begin{aligned} \sup _{x\in S} \left| E\left[ \widetilde{g_l} (x)\right] - E\left[ \widetilde{g_l}^* (x)\right] \right|= & {} \frac{1}{\widehat{\mathbf {n}}h^d}\left| E\left[ \displaystyle \sum _{{\mathbf {i}\in \mathcal {I}_\mathbf {n}}}Y_\mathbf{i}^{-l}{\mathbbm {1}}_{\{|Y_\mathbf{i}^{-1}|> {\mu }_\mathbf{n}\}}K_\mathbf{i}(x)\right] \right| \\\le & {} h^{-d} E\left[ |Y_\mathbf{1}^{-l}|{\mathbbm {1}}_{\{|Y_\mathbf{i}^{-1}|>{\mu }_\mathbf{n}\}}K_\mathbf{1}(x)\right] \\\le & {} h^{-d}E\left[ \exp \left( |Y_{1}^{-l}|/4\right) {\mathbbm {1}}_{\{|Y_\mathbf{i}^{-1}|>\mu _{n}\}}K_\mathbf{1}(x)\right] . \end{aligned}$$

  Furthermore, using The Holder’s inequality to show that,

  $$\begin{aligned} \sup _{x\in S}\left| E\left[ \widetilde{g_l} (x)\right] - E\left[ \widetilde{g_l}^* (x)\right] \right|\le & {} h^{-d} \left( E\left[ \exp \left( |Y_{1}^{-l}|/2\right) {\mathbbm {1}}_{\{Y_\mathbf{i}^{-1}|>{\mu }_{n}\}}\right] \right) ^{\frac{1}{2}} \left( E(K^{2}_\mathbf{1}(x))\right) ^{\frac{1}{2}}\\\le & {} h^{-d}\exp \left( -\mu _\mathbf{n}^l/4\right) \left( E\left[ \exp \left( |Y^{-l}|\right) \right] \right) ^{\frac{1}{2}}\left( E(K^{2}_\mathbf{1}(x)\right) ^{\frac{1}{2}} \\\le & {} C h^{\frac{-d}{2}}\exp \left( -{\mu }_\mathbf{n}^l/4\right) . \end{aligned}$$

  Since $\mu _\mathbf{n}=\widehat{\mathbf {n}}^{\gamma /2}$ then, we can write

  $$\begin{aligned} \sup _{x\in S} \left| E\left[ \widetilde{g_l} (x_k)\right] - E\left[ \widetilde{g_l}^* (x_k)\right] \right| =o\left( \left( \frac{ \log \widehat{\mathbf {n}}}{\widehat{\mathbf {n}}h^d}\right) ^{1/2}\right) . \end{aligned}$$

• Thirdly, the proof of the last claimed result (8) is based on the Markov’s inequality. Indeed, observe that, for all $\epsilon >0$

  $$\begin{aligned} {\mathbbm {P}}\left[ \sup _{x\in S}\left| \widetilde{g_l} (x)- \widetilde{g_l}^* (x)\right| >\epsilon \right]= & {} {\mathbbm {P}}\left( \frac{1}{\widehat{\mathbf {n}}h^d}\displaystyle \sum _{{\mathbf {i}\in \mathcal {I}_\mathbf {n}}}Y_\mathbf{i}^{-l}{\mathbbm {1}}_{|Y_\mathbf{i}^{-l}| >{\mu }_\mathbf{n}^l}K_{i}|>\epsilon \right) \\\le & {} \widehat{\mathbf {n}}{\mathbbm {P}}\left( \displaystyle |Y^{-l}|>{\mu }_\mathbf{n}\right) \\\le & {} \widehat{\mathbf {n}}\exp \left( -{\mu }_\mathbf{n}^l\right) E\left( \exp \left( |Y^{-1}|\right) \right) \\\le & {} C\widehat{\mathbf {n}}\exp \left( -{\mu }_\mathbf{n}^l\right) . \end{aligned}$$

  So,

  $$\begin{aligned} \displaystyle \sum _\mathbf{n}{\mathbbm {P}}\left( \sup _{x\in S}\left| \widetilde{g_l} (x)- \widetilde{g_l}^* (x)\right| >\epsilon _0\left( \sqrt{\frac{ \log \widehat{\mathbf {n}}}{\widehat{\mathbf {n}}h^d}}\right) \right) \le C\displaystyle \sum _\mathbf{n}\widehat{\mathbf {n}}\exp \left( -\mu _\mathbf{n}\right) . \end{aligned}$$

  (15)

  The use of the definition $\mu _\mathbf{n}$ complete the proof of Lemma. $\square$

Corollary 2

Under the hypotheses of Theorem 1, we obtain:

$$\begin{aligned} \sum _\mathbf{n}{\mathbbm {P}}\left( \inf _{x\in S}| \widetilde{g_2}(x)|\le \frac{\inf _{x\in S }g_2(x)}{2}\right) <\infty . \end{aligned}$$

Proof of Corollary 2

It is clear that

$$\begin{aligned} \inf _{x\in S } \widetilde{g_2}(x)< \frac{\inf _{x\in S }g_2(x)}{2}\Rightarrow \sup _{x\in S } |g_2(x)- \widetilde{g_2}(x)|> \frac{\inf _{x\in S }g_2(x)}{2}. \end{aligned}$$

Thus,

$$\begin{aligned} {\mathbbm {P}}\left( \inf _{x\in S}| \widetilde{g_2}(x)|\le \frac{\inf _{x\in S }g_2(x)}{2}\right)\le & {} {\mathbbm {P}}\left( \sup _{x\in S}|g_2(x)- \widetilde{g_2}(x)|\ge \frac{\inf _{x\in S }g_2(x)}{2}\right) . \end{aligned}$$

The use of the results of Lemma 1 and Lemma 2 complete the proof of the corollary. $\square$

Proof of Theorem 2 We write:

$$\begin{aligned} \widetilde{\theta }(x)- \theta (x)= \frac{1}{\widetilde{g_2}(x)} \left[ B_\mathbf{n}+V_\mathbf{n} \left( \widetilde{g_2}(x)-E\widetilde{g_2}(x)\right) \right] +V_\mathbf{n} \end{aligned}$$

where

$$\begin{aligned} V_\mathbf{n}=\frac{1}{E\widetilde{g_2}(x)g_2(x)}\left[ \Big [E\widetilde{g_1}(x)\Big ]g_2(x)-\Big [E\widetilde{g_2}(x)\Big ]g_1(x)\right] \end{aligned}$$

and

$$\begin{aligned} B_\mathbf{n}=\frac{1}{g_2(x)} \left[ \Big [\widetilde{g_1}(x)-E\widetilde{g_1}(x)\Big ]g_2(x)+\Big [E\widetilde{g_2}(x)-\widetilde{g_2}(x)\Big ]g_1(x)\right] . \end{aligned}$$

Therefore,

$$\begin{aligned} \widetilde{\theta }(x)- \theta (x)-V_\mathbf{n}= \frac{1}{\widetilde{g_2}(x)} \left[ B_\mathbf{n}+V_\mathbf{n} \left( \widetilde{g_2}(x)-E\widetilde{g_2}(x)\right) \right] \end{aligned}$$

(16)

Finally, Theorem 2 is a consequence of the following results (cf. Lemmas 3 and 4).

Lemma 3

Under the hypotheses of Theorem 2, we obtain:

$$\begin{aligned} \left( \frac{\widehat{\mathbf {n}}h^d}{g_2^2(x)\sigma ^2(x)}\right) ^{1/2}\left( B_\mathbf{n}-{\mathrm {E}}[B_\mathbf{n}]\right) \rightarrow N(0,1). \end{aligned}$$

Proof of Lemma 3

Considering the same notations of Lemma 1 and write,

$$\begin{aligned} B_\mathbf{n}=B_\mathbf{n}-B_\mathbf{n}^*+B_\mathbf{n}^* \end{aligned}$$

where

$$\begin{aligned} B_\mathbf{n}^*=\frac{1}{g_2(x)} \left[ \Big [\widetilde{g_1}^*(x)-E\widetilde{g_1}^*(x)\Big ]g_2(x)+\Big [E\widetilde{g_2}^*(x)-\widetilde{g_2}^*(x)\Big ]g_1(x)\right] . \end{aligned}$$

Similarly to Lemma 1 we get, for fixed $x\in {\mathbbm {R}}^d$

$$\begin{aligned} \left( \frac{\widehat{\mathbf {n}}h^d}{\left( g_2(x)\right) ^2 \sigma ^2(x)}\right) ^{1/2}\left| \widehat{g_l}(x)- \widehat{g_l}^*(x)\right| =o_p(1). \end{aligned}$$

As ${\mathrm {E}}[B_\mathbf{n}]={\mathrm {E}}[B_\mathbf{n}^*]=0$, then it suffices to show the asymptotic normality of

$$\begin{aligned} \left( \frac{\widehat{\mathbf {n}}h^d}{\left( g_2(x)\right) ^2 \sigma ^2(x)}\right) ^{1/2}\left| B_\mathbf{n}^*- E\left[ B_\mathbf{n}^*\right] \right| . \end{aligned}$$

For this, we put, for $\mathbf{i=1}\in I_\mathbf{n},$

$$\begin{aligned} \Lambda _{\mathbf {i}}:= & {} \frac{1}{\sqrt{h^d}}\left( K_i(Y_\mathbf{i}^{-1}g_2(x)-Y_\mathbf{i}^{-2}g_1(x)){\mathbbm {1}}_{|Y_\mathbf{i}^{-1}|<\mu _\mathbf{n} }\right. \nonumber \\&\left. -E\left[ K_i(Y_\mathbf{i}^{-1}g_2(x)-Y_\mathbf{i}^{-2}g_1(x)){\mathbbm {1}}_{|Y_\mathbf{i}^{-1}|<\mu _\mathbf{n} }\right] \right) . \end{aligned}$$

So,

$$\begin{aligned} \sqrt{\widehat{\mathbf {n}}h^d}\left[ \sigma _1(x)\right] ^{-1} \Big (B_\mathbf{n}^*-E\left[ B_\mathbf{n}^*\right] \Big )=\left( \widehat{\mathbf {n}}\sigma _1^2(x))\right) ^{-1/2}S_{\mathbf {n}} \end{aligned}$$

where $S_{\mathbf {n}}= \displaystyle \sum _{\mathbf{i}\in I_\mathbf{n}}\Lambda _\mathbf{i}$. Thus, our claimed result is, now

$$\begin{aligned} \left( \widehat{\mathbf {n}}\sigma _1^2(x))\right) ^{-1/2}S_{\mathbf {n}}\rightarrow \mathcal {N}(0,1). \end{aligned}$$

(17)

where $\sigma _1^2(x)=\left( g_2(x)\right) ^2 \sigma ^2(x)$. The proof of (17) follows the same lines of Lemma 3.2 in Tran (1990). It is based on spatial blocking technique for $S_{\mathbf {n}}=\displaystyle \sum _{\mathbf{i}\in I_\mathbf{n}}\Lambda _\mathbf{i}$.

Lemma 4

Under the hypotheses of Theorem 2, we obtain:

$$\begin{aligned} \widetilde{g}_2(x)\rightarrow g_2(x), \hbox { in probability,} \end{aligned}$$

$$\begin{aligned} \left( \frac{\widehat{\mathbf {n}}h^d}{g_2(x)^2\sigma ^2(x)}\right) ^{1/2}V_\mathbf{n} \rightarrow 0, \end{aligned}$$

and

$$\begin{aligned} \left( \frac{\widehat{\mathbf {n}}h^d}{g_2(x)^2\sigma ^2(x)}\right) ^{1/2}V_\mathbf{n} \left( \widetilde{g_2}(x)-E\widetilde{g_2}(x)\right) \rightarrow 0, \hbox {in probability}. \end{aligned}$$

Proof of Lemma 4 For the first limit, we have, by Lemma 1

$$\begin{aligned} E\left[ \widetilde{g}_2(x)-g_2(x))\right] \rightarrow 0 \end{aligned}$$

and by a similar argument as those used in the variance term in Lemma 2 we show that

$$\begin{aligned} Var\left[ \widetilde{g}_2(x)\right] \rightarrow 0 \end{aligned}$$

hence

$$\begin{aligned} \widetilde{g}_2(x)-g_2(x)\rightarrow 0\quad \text{ in } \text{ probability. } \end{aligned}$$

Next, it is clear that the second limit is consequence for the last convergence. So, it suffices to treat the last one. For this, we use the fact that

$$\begin{aligned} Var\left[ \left( \widetilde{g_2}(x)-E\widetilde{g_2}(x)\right) \right] = Var\left[ \widetilde{g}_2(x)\right] \rightarrow 0 \end{aligned}$$

and

$$\begin{aligned} V_\mathbf{n}=O(h^2) \quad \text{(see, } \text{ Lemma } \text{1) } \end{aligned}$$

The last part of Condition (H6$^{\prime }$) allows to deduce that

$$\begin{aligned} \left( \frac{\widehat{\mathbf {n}}h^d}{g_1(x)^2\sigma ^2(x)}\right) ^{1/2}V_\mathbf{n} \left( \widetilde{g_2}(x)-E\widetilde{g_2}(x)\right) )\rightarrow 0\quad \text{ in } \text{ probability. } \end{aligned}$$