Appendix
In what follows, when no confusion is possible, we will denote by C and \(C'\) some strictly positive generic constants. Moreover, we will denote by
$$\begin{aligned} \displaystyle K_i=K(h^{-1}\varrho (x,X_i)) \hbox { and } \beta _i=\beta (x, X_i) \hbox { for } i=1,\ldots ,n. \end{aligned}$$
Then, in order to establish our asymptotic results the following lemmas will be needed.
Lemma 1
Let \((V_n)\) be a sequence of vectorial functions, such that
-
(i)
for all \(\lambda \ge 1\) and all vector \(\delta \)
$$\begin{aligned} ^t(\delta ) V_n(\lambda \delta )\le \ ^t(\delta ) V_n (\delta ), \end{aligned}$$
where \(^t u\) denotes the transpose of the vector u.
-
(ii)
for any positive definite matrix D and vectorial sequence \((A_n)\), such that \(\displaystyle \mathbb {P}\left( \Vert A_n\Vert \ge A\right) \rightarrow 0\) for some \(A>0\), we have
$$\begin{aligned} \displaystyle \sup _{ \Vert \delta \Vert \le M}\Vert V_n(\delta )+\lambda _0D\delta -A_n\Vert =o_{p}(1)\ \hbox { for } \ \frac{A}{\lambda _0\lambda _1(D)}< M<\infty , \end{aligned}$$
with \(\lambda _1(D)\) is the minimal eigenvalue of D and \(\lambda _0>0\).
then, for any vectorial sequence \((\delta _n)\), such that \(V_n(\delta _n)=o_{p}(1)\), we have
$$\begin{aligned} \Vert \delta _n\Vert \le M, \hbox { in probability.} \end{aligned}$$
\(\Vert A_n\Vert =o_{a.co .}(1)\) and \(\displaystyle \sup _{ \Vert \delta \Vert \le M}\Vert V_n(\delta )+\lambda _0D\delta -A_n\Vert =o_{a.co.}(1),\) then, for any vectorial sequence \((\delta _n)\) such that \(\displaystyle V_n(\delta _n)=o_{a.co.}(1)\), we have
$$\begin{aligned} \Vert \delta _n\Vert \le M, \hbox { a.co.} \end{aligned}$$
Proof of Lemma 1
Notice that, proofs of both results are similar. They are based on same arguments as in Koenker and Zhao (1996). For a sake of shortness, we prove the second case which is more general. Indeed, for \(\eta >0\), we have:
$$\begin{aligned} \mathbb {P}\left( \Vert \delta _n\Vert \ge M\right)= & {} \mathbb {P}\left( \Vert \delta _n\Vert \ge M,\Vert V_n(\delta _n)\Vert<\eta \right) + \mathbb {P}\left( \Vert V_n(\delta _n)\Vert \ge \eta \right) \\\le & {} \mathbb {P}\left( \inf _{\Vert \delta \Vert \ge M}\Vert V_n(\delta )\Vert <\eta \right) + \mathbb {P}\left( \Vert V_n(\delta _n)\Vert \ge \eta \right) . \end{aligned}$$
Since, for any \(\eta >0\), \(\displaystyle \sum _n\mathbb {P}\left( \Vert V_n(\delta _n)\Vert \ge \eta \right) <\infty \), then, all it remains to show is that
$$\begin{aligned} \sum _n\mathbb {P}\left( \inf _{\Vert \delta \Vert \ge M}\Vert V_n(\delta )\Vert<\eta \right) <\infty . \end{aligned}$$
Now, it is clear that any vector \(\delta \) such that \(\Vert \delta \Vert \ge M \), can be written as \(\delta =\lambda \delta _1\) for \(\lambda \ge 1\) and \(\Vert \delta _1\Vert =M\). So, by condition (i) we get
$$\begin{aligned} \left\| V_n(\delta )\right\| =\left\| -\frac{^t\delta _1V_n(\delta )}{M}\right\| \ge -\frac{^t\delta _1V_n(\delta )}{M}\ge -\frac{^t\delta _1V_n(\delta _1)}{M}, \end{aligned}$$
which implies that
$$\begin{aligned} \mathbb {P}\left( \inf _{\Vert \delta \Vert \ge M}\Vert V_n(\delta )\Vert<\eta \right) \le \mathbb {P}\left( \inf _{\Vert \delta _1\Vert =M }\left[ -\frac{\delta _1V_n(\delta _1)}{M}\right] <\eta \right) . \end{aligned}$$
Thus, it suffices to evaluate this last quantity. To do that, we write
$$\begin{aligned}&\mathbb {P}\left( \inf _{\Vert \delta _1\Vert =M}\left[ -\frac{^t\delta _1V_n(\delta _1)}{M}\right]<\eta \right) \\&\quad \le \mathbb {P}\left( \inf _{\Vert \delta _1\Vert =M}\left[ -^t\delta _1V_n(\delta _1)\right] <\eta M, \right. \\&\qquad \left. \displaystyle \inf _{\Vert \delta _1\Vert =M }\left[ -^t\delta _1\left( -f^x(t_\alpha (x))D\delta _1+A_n\right) \right] \ge 2\eta M \right) \\&\qquad +\,\mathbb {P}\left( \inf _{\Vert \delta _1\Vert =M }\left[ -^t\delta _1\left( -\lambda _0D \delta _1+A_n\right) \right] \le 2\eta M \right) \\&\quad \le \mathbb {P}\left( \sup _{\Vert \delta _1\Vert =M}\left\| V_n(\delta _1)+\lambda _0D\delta _1-A_n\right\| \ge \eta \right) \\&\qquad +\,\mathbb {P}\left( \Vert A_n\Vert \ge \lambda _0\lambda _1(D)M -2\eta \right) . \end{aligned}$$
Finally, under conditions (ii), we can choose \(\eta \) for which
$$\begin{aligned} \sum _n\mathbb {P}\left( \sup _{\Vert \delta \Vert =M }\left\| V_n(\delta _1)+\lambda _0D\delta _1-A_n\right\| \ge \eta \right) <\infty , \end{aligned}$$
and
$$\begin{aligned} \sum _n \mathbb {P}\left( \Vert A_n\Vert \ge \lambda _0\lambda _1(D)M -2\eta \right) <\infty . \end{aligned}$$
It follows that
$$\begin{aligned} \sum _n\mathbb {P}\left( \Vert \delta _n\Vert \ge M\right) <\infty . \end{aligned}$$
\(\square \)
Proof of Theorem 1
We Define, for all \(\displaystyle \delta = \left( \begin{array}{c}c \\ d \\ \end{array}\right) \),
$$\begin{aligned} \psi _\alpha (\delta )=\alpha -\mathbb {1}_{\displaystyle \left[ Y_i\le (c+a)+(h_n^{-1}d+b)\beta _i\right] ,} \end{aligned}$$
and we consider the following vectorial sequence
$$\begin{aligned} V_n(\delta )=\frac{1}{n\phi _x(h)}\sum _{i=1}^n\psi _\alpha (\delta ) \left( \begin{array}{c} 1 \\ h^{-1}\beta _i \\ \end{array} \right) K_i, \end{aligned}$$
and
$$\begin{aligned} A_n=V_n (\delta _0)\quad \text{ with }\quad \delta _0=\left( \begin{array}{c} 0 \\ 0 \\ \end{array} \right) . \end{aligned}$$
Then, the proof of the first result of Theorem 1 is based on the application of the second part of Lemma 1 on the sequences \((V_n, A_n)\) with \(\delta _n\displaystyle =\left( \begin{array}{c} \widehat{a}-a \\ h(\widehat{b}-b) \end{array} \right) \). Furthermore, the second result of Theorem 1 may be obtained by applying the first part of Lemma 1 to the sequences
$$\begin{aligned} V_n'(\delta )=\frac{\sqrt{n\phi _x(h)}}{n\phi _x(h)}\sum _{i=1}^n\Psi _\alpha (\delta ) \left( \begin{array}{c} 1 \\ h^{-1}\beta _i \\ \end{array} \right) K_i, \; A_n'=V_n'(\delta _0) \quad \text{ and }\quad \delta _n'=\sqrt{n\phi _x(h)}\delta _n, \end{aligned}$$
where
$$\begin{aligned} \Psi _\alpha (\delta )=\alpha -\mathbb {1}_{\displaystyle \left[ Y_i\le \left( \frac{1}{\sqrt{n\phi _x(h)}}c+a\right) +\left( \frac{1}{h\sqrt{n\phi _x(h)}}d+b\right) \beta _i\right] }. \end{aligned}$$
\(\square \)
Since \(\mathbb {1}_{\displaystyle [Y_i\le \cdot ]} \) is a monotone and increasing function then the condition (i), of Lemma 1, holds. So, the first result of Theorem 1 is a consequence of the following lemmas.
Lemma 2
Under Assumptions (H1)–(H5), we have
$$\begin{aligned} \Vert A_n\Vert =O(h^{\min (k_1,k_2)})+O_{a.co.}\left( \frac{\log n}{n\,\phi _x(h)}\right) ^{1/2}. \end{aligned}$$
Lemma 3
Under Assumptions (H1)–(H5), we have
$$\begin{aligned} \displaystyle \sup _{ \Vert \delta \Vert \le M}\Vert V_n(\delta )+\lambda _0D\delta -A_n\Vert =o_{a.co.}(1), \end{aligned}$$
with
$$\begin{aligned} D=\left( \begin{array}{ll} K(1)-\int _{-1}^{1}K'(t)\chi _x(t)dt&{} K(1)-\int _{-1}^{1}(tK(t))'\chi _x(t)dt \\ \\ K(1)-\int _{-1}^{1}(tK(t))'\chi _x(t)dt&{} K(1)-\int _{-1}^{1}(t^2K(t))'\chi _x(t)dt \end{array}\right) , \end{aligned}$$
and
$$\begin{aligned} \lambda _0=f^x(t_\alpha (x)). \end{aligned}$$
Concerning the asymptotic normality result we use the following lemmas.
Lemma 4
Under assumptions (H1)–(H5), we have
$$\begin{aligned} \Vert A_n'\Vert =O_p\left( 1\right) . \end{aligned}$$
Lemma 5
Under assumptions (H1)–(H5), we have
$$\begin{aligned} \displaystyle \sup _{ \Vert \delta '\Vert \le M}\Vert V_n'(\delta )+f^x(t_\alpha (x))D\delta '-A_n'\Vert =o_p(1). \end{aligned}$$
Corollary 1
Under assumptions of Lemma 5, we have
$$\begin{aligned}&\left( \sqrt{n\phi _x(h)}\right) \left( \widehat{t_\alpha }(x)-t_\alpha (x)\right) \\&\quad =\frac{1}{f^x(t_\alpha (x))(a_1a_3-a_2^2)\sqrt{n\phi _x(h)}}\sum _{i=1}^n\left( \alpha -\mathbb {1}_{\displaystyle [Y_i\le a+b\beta _i]}\right) \left( a_3K_i-a_2h^{-1}\beta _iK_i\right) . \end{aligned}$$
Lemma 6
Under the hypotheses of Theorem 1, we have, for all \(u\in \mathbb {R}\),
$$\begin{aligned} \sqrt{n\phi _x(h)}\left[ f^x(t_\alpha (x))\alpha (1-\alpha )\left( a_3^2D_1-2a_2a_3D_2+a_2^2D_3\right) \right] ^{-1}\left( \widehat{\Phi }(x)-\mathbb {E}\left[ \widehat{\Phi }(x)\right] \right) \end{aligned}$$
converges, in distribution, toward \({\mathcal {N}}(0,1)\)\(\hbox { as } n\rightarrow \infty \), where
$$\begin{aligned} \widehat{\Phi }(x)=\frac{1}{n\phi _x(h)}\sum _{i=1}^n\left( \alpha -\mathbb {1}_{\displaystyle [Y_i\le a+b\beta _i]}\right) \left( a_3K_i-a_2h^{-1}\beta _iK_i\right) . \end{aligned}$$
Proof of Lemma 2
Firstly, we set
$$\begin{aligned} \Delta ^1_i= \left( \alpha -\mathbb {1}_{\displaystyle [Y_i\le a+b\beta _i]}\right) K_i -\mathbb {E}\left[ \left( \alpha -\mathbb {1}_{\displaystyle [Y_i\le a+b\beta _i]}\right) K_i\right] \end{aligned}$$
and
$$\begin{aligned} \Delta ^2_i= \left( \alpha -\mathbb {1}_{\displaystyle [Y_i\le a+b\beta _i]}\right) \beta _iK_i -\mathbb {E}\left[ \left( \alpha -\mathbb {1}_{\displaystyle [Y_i\le a+b\beta _i]}\right) K_i\right] . \end{aligned}$$
Then
$$\begin{aligned} A_n-\mathbb {E}[A_n]= \left( \begin{array}{c} A_n^1 \\ A_n^2 \\ \end{array} \right) \end{aligned}$$
where
$$\begin{aligned} \left\{ \begin{array}{c} A_n^1=\displaystyle \frac{1}{n\phi _x(h)}\sum _{i=1}^n\Delta ^1_i \\ A_n^2=\displaystyle \frac{1}{nh\phi _x(h)}\sum _{i=1}^n\Delta ^2_i. \end{array} \right. \end{aligned}$$
It is clear that
$$\begin{aligned} |\Delta ^1_i|\le C \quad \text{ and }\quad |\Delta ^2_i|\le C' h . \end{aligned}$$
Moreover,
$$\begin{aligned} \mathbb {E}\left[ \Delta ^1_i\right] ^2\le C\phi _x(h) \quad \text{ and }\quad \mathbb {E}\left[ \Delta ^2_i\right] ^2\le C' h^2 \phi _x(h) . \end{aligned}$$
It follows that
$$\begin{aligned} A_n^1-\mathbb {E}\left[ A_n^1\right] =O_{a.co.}\left( \sqrt{\frac{\log n}{n\phi _x(h)}}\right) \end{aligned}$$
and
$$\begin{aligned} A_n^2-\mathbb {E}\left[ A_n^2\right] =O_{a.co.}\left( \sqrt{\frac{\log n}{n\phi _x(h)}}\right) . \end{aligned}$$
On the other hand, we have
$$\begin{aligned} \mathbb {E}\left[ A_n^1\right]= & {} \frac{1}{\phi _x(h) } \mathbb {E}\left[ \left( \alpha -\mathbb {1}_{\displaystyle [Y_1\le a+b\beta _1]}\right) K_1\right] \\\le & {} \displaystyle \frac{1}{\phi _x(h) }\mathbb {E}\left| F^x(t_\alpha (x))-F^{X}(a+b\beta _1)K_1\right| \\= & {} O (h^{\min (k_1,k_2)}). \end{aligned}$$
Similarly,
$$\begin{aligned} \mathbb {E}\left[ A_n^2\right]= & {} \frac{1}{h\phi _x(h) } \mathbb {E}\left[ \left( \alpha -\mathbb {1}_{\displaystyle [Y_1\le a+b\beta _1]}\right) \beta _1K_1\right] \\\le & {} \displaystyle \frac{1}{h\phi _x(h) }\mathbb {E}\left| F^x(t_\alpha (x))-F^{X}(a+b\beta _1)\beta _1K_1\right| \\= & {} O (h^{\min (k_1,k_2)}). \end{aligned}$$
Therefore,
$$\begin{aligned} \Vert A_n\Vert = O( h^{\min (k_1,k_2)})+O_{a.co.}\left( \frac{\log n}{n\,\phi _x(h)}\right) ^{1/2}. \end{aligned}$$
\(\square \)
Proof of Lemma 3
We prove that
$$\begin{aligned} \sup _{\Vert \delta \Vert \le M} \Vert V_n(\delta )-A_n-\mathbb {E}\left[ V_n(\delta )-A_n\right] \Vert =O_{a.co.}\left( \sqrt{\frac{\log n}{n\phi _x(h)}}\right) , \end{aligned}$$
(2)
and
$$\begin{aligned} \sup _{\Vert \delta \Vert \le M} \Vert \mathbb {E}\left[ V_n(\delta )-A_n\right] +f^x(t_\alpha (x))D\delta \Vert =O(h^{\min (k_1,k_2)}). \end{aligned}$$
(3)
In order to show (2) we use the compactness property of the ball B(0, M) in \(\mathbb {R}^2\) and we write
$$\begin{aligned} B(0,M)\subset \bigcup _{j=1}^{d_n}B(\delta _j,l_n), \ \delta _j=\left( \begin{array}{c} c_j \\ d_j \\ \end{array} \right) \hbox { and } l_n=d_n^{-1}=1/\sqrt{n}. \end{aligned}$$
Then, we take \(j(\delta )=\arg \min _j|\delta -\delta _j|\) and we use the fact that
$$\begin{aligned}&\sup _{\Vert \delta \Vert \le M} \Vert V_n(\delta )-A_n-\mathbb {E}\left[ V_n(\delta )-A_n\right] \Vert \\&\quad \le \sup _{\Vert \delta \Vert \le M}\Vert V_n(\delta )-V_n(\delta _j)\Vert \\&\qquad +\, \sup _{\Vert \delta \Vert \le M}\Vert V_n(\delta _j)-A_n-\mathbb {E}\left[ V_n(\delta _j)-A_n\right] \Vert \\&\qquad +\,\sup _{\Vert \delta \Vert \le M}\Vert \mathbb {E}\left[ V_n(\delta )-V_n(\delta _j)\right] . \end{aligned}$$
Since \(|\mathbb {1}_{[Y<a]}-\mathbb {1}_{[Y<b]}| \le \mathbb {1}_{\left[ |Y-b|\le |a-b|\right] } \) then
$$\begin{aligned} \sup _{\Vert \delta \Vert \le M} \Vert V_n(\delta )-V_n(\delta _j)\Vert \le \frac{1}{n\phi _x(h)}\sum _i Z_i , \end{aligned}$$
where
$$\begin{aligned} Z_i= \sup _{\Vert \delta \Vert \le M}\mathbb {1}_{\left[ |Y_i-(c_j+a)-(h^{-1}d_j+d)\beta _i|\le Cl_n\right] }\left\| \left( \begin{array}{c} 1 \\ h^{-1}\beta _i \\ \end{array} \right) \right\| K_i . \end{aligned}$$
It is clear that
$$\begin{aligned} |Z_i|\le C,\quad \mathbb {E}[Z_i]=O( l_n\phi _x(h)) \quad \text{ and } \quad \mathbb {E}[Z_i^2]=O(l_n\phi _x(h)). \end{aligned}$$
By using the fact that
$$\begin{aligned} l_n=o\left( \frac{\log n}{n\phi _x(h)}\right) ^{1/2}, \end{aligned}$$
(4)
we get
$$\begin{aligned} \sup _{\Vert \delta \Vert \le M} \Vert V_n(\delta )-V_n(\delta _j)\Vert =O_{a.co.}\left( \frac{\log n}{n\phi _x(h)}\right) ^{1/2}. \end{aligned}$$
Concerning the last term, we have
$$\begin{aligned} \sup _{\Vert \delta \Vert \le M}\Vert \mathbb {E}\left[ V_n(\delta )-V_n(\delta _j)\right] \Vert \le \frac{1}{\phi _x(h)}\mathbb {E}[Z_1]\le Cl_n, \end{aligned}$$
then, by (4) we obtain
$$\begin{aligned} \sup _{\Vert \delta \Vert \le M} \Vert \mathbb {E}\left[ V_n(\delta )-V_n(\delta _j)\right] \Vert =o_{a.co.}\left( \frac{\log n}{n\phi _x(h)}\right) ^{1/2}. \end{aligned}$$
Now, we are dealing with the quantity
$$\begin{aligned} \sup _{\Vert \delta \Vert \le M}\Vert V_n(\delta _j)-A_n-\mathbb {E}\left[ V_n(\delta _j)-A_n\right] \Vert . \end{aligned}$$
For this, we set
$$\begin{aligned} V_n(\delta _j)-A_n-\mathbb {E}\left[ V_n(\delta _j)-A_n\right] = \left( \begin{array}{c} \displaystyle W_n^1(\delta _j) \\ \displaystyle W_n^2(\delta _j) \\ \end{array} \right) , \end{aligned}$$
where
$$\begin{aligned} \left\{ \begin{array}{l} W_n^1(\delta _j)=\displaystyle \frac{1}{n\phi _x(h)}\sum _{i=1}^n\Gamma ^1_i \\ W_n^2(\delta _j)=\displaystyle \frac{1}{nh\phi _x(h)}\sum _{i=1}^n\Gamma ^2_i \end{array} \right. \end{aligned}$$
with
$$\begin{aligned} \Gamma ^1_i=\left( \psi _\alpha (\delta _j)-\psi _\alpha (\delta _0)\right) K_i-\mathbb {E}\left[ \left( \psi _\alpha (\delta _j)-\psi _\alpha (\delta _0)\right) K_i\right] , \end{aligned}$$
and
$$\begin{aligned} \Gamma ^2_i= \left( \psi _\alpha (\delta _j)-\psi _\alpha (\delta _0)\right) \beta _iK_i-\mathbb {E}\left[ \left( \psi _\alpha (\delta _j)-\psi _\alpha (\delta _0) \right) \beta _iK_i\right] . \end{aligned}$$
It is clear that
$$\begin{aligned} |\Gamma ^1_i|\le C \quad \text{ and }\quad |\Gamma ^2_i|\le C' h . \end{aligned}$$
Moreover
$$\begin{aligned} \mathbb {E}\left[ \Gamma ^1_i\right] ^2\le C\phi _x(h) \quad \text{ and }\quad \mathbb {E}\left[ \Gamma ^2_i\right] ^2\le C' h^2 \phi _x(h). \end{aligned}$$
It follows that, there exits \(\eta >0\) such that
$$\begin{aligned}&\sum _n\mathbb {P}\left( \sup _{\Vert \delta \Vert \le M}\Vert V_n(\delta _j)-A_n-\mathbb {E}\left[ V_n(\delta _j)-A_n\right] \Vert \ge \eta \sqrt{\frac{\log n}{n\phi _x(h)}} \right) \\&\quad \le \sum _n d_n \max _j \mathbb {P}\left( \Vert V_n(\delta _j)-A_n-\mathbb {E}\left[ V_n(\delta _j)-A_n\right] \Vert \ge \eta \sqrt{\frac{\log n}{n\phi _x(h)}} \right) <\infty , \end{aligned}$$
which complete the proof of (2).
Concerning the term (3) we use same arguments as for treating (2). In fact, we write
$$\begin{aligned} V_n(\delta )-A_n= \left( \begin{array}{c} W_n^1(\delta ) \\ W_n^2(\delta ) \end{array} \right) , \end{aligned}$$
where
$$\begin{aligned} W_n^1(\delta )=\frac{1}{n\phi _x(h)}\sum _{i=1}^n\left( \psi _\alpha (\delta )-\psi _\alpha (\delta _0)\right) K_i, \end{aligned}$$
and
$$\begin{aligned} W_n^2(\delta )=\frac{1}{nh\phi _x(h)}\sum _{i=1}^n \left( \psi _\alpha (\delta )-\psi _\alpha (\delta _0)\right) \beta _iK_i. \end{aligned}$$
On the other hand, we have
$$\begin{aligned} \mathbb {E}\left[ W_n^1(\delta )\right]= & {} -\frac{1}{\phi _x(h) }\mathbb {E}\left[ \left( \mathbb {1}_{\left[ Y_1\le (c+a)+(h^{-1}d+b)\beta _1\right] }-\mathbb {1}_{\left[ Y_1\le a+b\beta _1\right] }\right) K_i\right] \\= & {} -\frac{1}{\phi _x(h) }\mathbb {E}\left[ \left( F^X(((c+a)+(h^{-1}d+b)\beta _1)-F^X(a+b\beta _1)\right) K_1\right] \\= & {} -\frac{1}{\phi _x(h) }\mathbb {E}\left[ \left( F^x(((c+a)+(h^{-1}d+b)\beta _1)-F^x(a+b\beta _1)\right) K_1\right] +O(h^{k_1})\\= & {} -\frac{1}{\phi _x(h) }\mathbb {E}\left[ f^x(a+b\beta _1)\left( 1, h^{-1} \beta _1 \right) \delta K_1\right] +O(h^{k_1})+o\left( \Vert \delta \Vert \right) \\= & {} - f^x(t_\alpha (x))\frac{1}{\phi _x(h) }\left( \mathbb {E}K_1, h^{-1}\mathbb {E}[\beta _1 K_1] \right) \delta +O(h^{\min (k_1,k_2)})+o\left( \Vert \delta \Vert \right) . \end{aligned}$$
Similarly
$$\begin{aligned} \mathbb {E}\left[ W_n^2(\delta )\right]= & {} -f^x(t_\alpha (x))\frac{1}{h\phi _x(h) }\left( \mathbb {E}\beta _1K_1, h^{-1}\mathbb {E}[\beta _1^2 K_1] \right) \delta \\&+O(h^{\min (k_1,k_2)})+o\left( \Vert \delta \Vert \right) . \end{aligned}$$
It follows that
$$\begin{aligned} \mathbb {E}\left[ V_n(\delta )-A_n\right]= & {} -f^x(t_\alpha (x))\frac{1}{\phi _x(h) }\left( \begin{array}{ll} \mathbb {E}\left[ K_i\right] &{} \mathbb {E}\left[ K_ih^{-1}\beta _i\right] \\ \mathbb {E}\left[ K_ih^{-1}\beta _i\right] &{} \mathbb {E}\left[ h^{-2}\beta _i^2K_i\right] \\ \end{array} \right) \delta \\&+O(h^{\min (k_1,k_2)})+o\left( \Vert \delta \Vert \right) . \end{aligned}$$
Using the same ideas as in Rachdi et al. (2014) we show that
$$\begin{aligned} h^{-a}\mathbb {E}[\beta ^{-a}K_i^b]=\phi _x(h)\left( K^b(1)-\int _{-1}^{1} (u^aK^b(u))'\chi _x(u)du\right) +o(\phi _x(h)). \end{aligned}$$
Hence, we can conclude that
$$\begin{aligned} \sup _{\Vert \delta \Vert \le M} \Vert \mathbb {E}\left[ V_n(\delta )-A_n\right] +f^x(t_\alpha (x))D\delta + o(\Vert \delta \Vert )\Vert =O(h^{\min (k_1,k_2)}), \end{aligned}$$
which implies the result (3). \(\square \)
Proof of Lemma 4
It suffices to prove that
$$\begin{aligned} n\phi _x(h)\ var[c_1A_n^1+c_2A_n^2]\rightarrow C, \hbox { for } \left( \begin{array}{cc} c_1 \\ c_2 \\ \end{array} \right) \in \mathbb {R}^2, \end{aligned}$$
where \(A_n^1\) and \(A_n^2\) are defined in the Proof of Lemma 2 and C is an arbitrary constant. Indeed, by the stationarity property we write
$$\begin{aligned} n\phi _x(h)\ var[c_1A_n^1+c_2A_n^2]= & {} \frac{c_1^2}{\phi _x(h)} \ var[\Delta _1^1]+ \frac{c_2^2}{h^2\phi _x(h)} \ var[\Delta _1^2]\\&+\,2\frac{c_1c_2}{h\phi _x(h)}\ cov(\Delta _1^1,\Delta _1^2), \end{aligned}$$
and by some simple calculations we get
$$\begin{aligned} \frac{1}{\phi _x(h)}\ var[\Delta _1^1]\rightarrow & {} \alpha (1-\alpha )\left( K^2(1)- \int _{-1}^{1}(K^2(t))'\chi _x(t)dt\right) , \\ \frac{1}{h^2\phi _x(h)}\ var[\Delta _1^2]\rightarrow & {} \alpha (1-\alpha )\left( K^2(1)-\int _{-1}^{1}(t^2K^2(t))'\chi _x(t)dt\right) , \end{aligned}$$
and
$$\begin{aligned} \frac{c_1c_2}{h\phi _x(h)}\ cov(\Delta _1^1,\Delta _1^2)\rightarrow \alpha (1-\alpha )\left( K^2(1)-\int _{-1}^{1}(tK^2(t))'\chi _x(t)dt\right) , \end{aligned}$$
which yields to the proof of this lemma. \(\square \)
Proof of Lemma 5
The proof is based on the same arguments as in the Proof of Lemma 3. Precisely, it is based on the following two results
$$\begin{aligned}&\sup _{\Vert \delta \Vert \le M} \Vert \mathbb {E}\left[ V_n'(\delta )-A_n'\right] +f^x(t_\alpha (x))D\delta \Vert \nonumber \\&\quad =O \left( \sqrt{n\phi _x(h)}h^{\min (k_1,k_2)}\right) =o(1), \end{aligned}$$
(5)
and
$$\begin{aligned} \sup _{\Vert \delta \Vert \le M} \Vert V_n(\delta )-A_n-\mathbb {E}\left[ V_n(\delta )-A_n\right] \Vert =o_{p}\left( 1\right) . \end{aligned}$$
(6)
For (5), we write
$$\begin{aligned} V_n'(\delta )-A_n'= \sqrt{n\phi _x(h)}\left( \begin{array}{c} W_n^{1'}(\delta ) \\ W_n^{2'}(\delta ) \\ \end{array} \right) , \end{aligned}$$
where
$$\begin{aligned} W_n^{1'}(\delta )=\frac{1}{n\phi _x(h)}\sum _{i=1}^n\left( \Psi _\alpha (\delta )-\Psi _\alpha (\delta _0)\right) K_i, \end{aligned}$$
and
$$\begin{aligned} W_n^{2'}(\delta )=\frac{1}{nh\phi _x(h)}\sum _{i=1}^n\left( \Psi _\alpha (\delta )-\Psi _\alpha (\delta _0)\right) \beta _iK_i. \end{aligned}$$
Therefore
$$\begin{aligned}&\mathbb {E}\left[ V_n'(\delta )-A_n'\right] \\&\quad = -f^x(t_\alpha (x))\frac{1}{\phi _x(h) }\left( \begin{array}{ll} \mathbb {E}\left[ K_i\right] &{} \mathbb {E}\left[ K_ih^{-1}\beta _i\right] \\ \mathbb {E}\left[ K_ih^{-1}\beta _i\right] &{} \mathbb {E}\left[ h^{-2}\beta _i^2K_i\right] \\ \end{array} \right) \delta +O(h^{\min (k_1,k_2)})+o\left( \Vert \delta \Vert \right) . \end{aligned}$$
Now, for the term (6), we keep notations of Lemma 3 with other choices of \(l_n\) and \(d_n\). Here, we set \(l_n=d_n^{-1}= O( 1/\log n)\) and we write
$$\begin{aligned}&\sup _{\Vert \delta \Vert \le M} \Vert V_n'(\delta )-A_n'-\mathbb {E}\left[ V_n'(\delta )-A_n'\right] \Vert \\&\quad \le \sup _{\Vert \delta \Vert \le M} \Vert V_n(\delta )-V_n(\delta _j)\Vert \\&\qquad +\, \sup _{\Vert \delta \Vert \le M}\Vert V_n'(\delta _j)-A_n'-\mathbb {E}\left[ V_n'(\delta _j)-A_n'\right] \Vert \\&\qquad + \,\sup _{\Vert \delta \Vert \le M}\Vert \mathbb {E}\left[ V_n'(\delta )-V_n'(\delta _j)\right] . \end{aligned}$$
Once again, we write
$$\begin{aligned} \sup _{\Vert \delta \Vert \le M} \Vert V_n'(\delta )-V_n'(\delta _j)\Vert \le \frac{\sqrt{n\phi _x(h)}}{n\phi _x(h)}\sum _i Z_i, \end{aligned}$$
where
$$\begin{aligned} Z_i= \sup _{\Vert \delta \Vert \le M}\mathbb {1}_{\left[ \left| Y_i-\left( \frac{1}{\sqrt{n\phi _x(h)}}c_j+a\right) -\left( \frac{1}{h\sqrt{n\phi _x(h)}}d_j+d\right) \beta _i\right| \le C\frac{l_n}{\sqrt{n\phi _x(h)}}\right] }\left\| \left( \begin{array}{c} 1 \\ h^{-1}\beta _i \\ \end{array} \right) \right\| K_i . \end{aligned}$$
It is clear that
$$\begin{aligned} \mathbb {E}[Z_i]=O( n^{-1/2}l_n\sqrt{\phi _x(h)}) \quad \text{ and } \quad \ var[Z_i]=O(n^{-1/2}l_n\sqrt{\phi _x(h)}). \end{aligned}$$
Thus
$$\begin{aligned} \sup _{\Vert \delta \Vert \le M} \Vert V_n'(\delta )-V_n'(\delta _j)\Vert =o_{p}\left( 1\right) . \end{aligned}$$
(7)
Concerning the last term we have
$$\begin{aligned} \sup _{\Vert \delta \Vert \le M}\Vert \mathbb {E}\left[ V_n'(\delta )-V_n'(\delta _j)\right] \le Cl_n=o(1). \end{aligned}$$
(8)
Now, we treat the second term. To do that we write
$$\begin{aligned} V_n'(\delta _j)-A_n'-\mathbb {E}\left[ V_n'(\delta _j)-A_n'\right] = \left( \begin{array}{c} \sqrt{n\phi _x(h)} U_n^1(\delta _j) \\ \sqrt{n\phi _x(h)} U_n^2(\delta _j) \\ \end{array} \right) , \end{aligned}$$
where
$$\begin{aligned} U_n^1(\delta _j)=\frac{1}{n\phi _x(h)}\sum _{i=1}^n\Gamma ^{1'}_i \hbox { and } U_n^2(\delta _j)=\frac{1}{nh\phi _x(h)}\sum _{i=1}^n\Gamma ^{2'}_i, \end{aligned}$$
with
$$\begin{aligned} \Gamma ^{1'}_i=\left( \Psi _\alpha (\delta _j)-\Psi _\alpha (\delta _0)\right) K_i-\mathbb {E}\left[ \left( \Psi _\alpha (\delta _j)-\Psi _\alpha (\delta _0)\right) K_i \right] , \end{aligned}$$
and
$$\begin{aligned} \Gamma ^{2'}_i= \left( \Psi _\alpha (\delta _j)-\Psi _\alpha (\delta _0)\right) \beta _iK_i-\mathbb {E}\left[ \left( \Psi _\alpha (\delta _j)-\Psi _\alpha (\delta _0)\right) \beta _iK_i\right] . \end{aligned}$$
It is clear that
$$\begin{aligned} \ var\left[ \Gamma ^{1'}_i\right] =O(n^{-1/2}\sqrt{\phi _x(h)}) \quad \text{ and }\quad \ var\left[ \Gamma ^{2'}_i\right] =O(h^2n^{-1/2}\sqrt{\phi _x(h)}) . \end{aligned}$$
It follows that
$$\begin{aligned} \displaystyle \ var| \sqrt{n\phi _x(h)}\left( U_n^1(\delta _j)-\mathbb {E}\left[ U_n^1(\delta _j)\right] \right) =O\left( \frac{1 }{\sqrt{n\phi _x(h)}}\right) , \end{aligned}$$
(9)
and
$$\begin{aligned} \displaystyle \ var| \sqrt{n\phi _x(h)}\left( U_n^2(\delta _j)-\mathbb {E}\left[ U_n^2(\delta _j)\right] \right) =O\left( \frac{1 }{\sqrt{n\phi _x(h)}}\right) . \end{aligned}$$
(10)
Thus, the proof of this lemma is a consequence of the Assumption (H5) and (7)–(10). \(\square \)
Proof of Corollary 1
Since \(\Vert \delta _n'\Vert \le M\) and \(V_n'(\delta _n')=0\) then, from Lemma 5, we have
$$\begin{aligned} f^x(t_\alpha (x))D\delta _n'=A_n'+o_p(1), \end{aligned}$$
which implies that
$$\begin{aligned} \delta _n'=\frac{1}{f^x(t_\alpha (x))}D^{-1}A_n'+ o_p(1). \end{aligned}$$
Now, it suffices to replace \(A_n'\) and \(D^{-1}\) by their expressions to achieve the proof of this corollary. \(\square \)
Proof of Lemma 6
We use a Liapounov’s Theorem (see Loève 1963, p. 275). In the first step we consider the asymptotic behavior of the variance term
$$\begin{aligned} n\phi _x(h)\ var\left[ \widehat{\Phi }(x)\right] . \end{aligned}$$
From Lemma 4, we have
$$\begin{aligned} n\phi _x(h)\ var\left[ \widehat{\Phi }(x)\right] \rightarrow \alpha (1-\alpha ) a_3^2D_1-2a_2a_3D_2+a_2^2D_3 \hbox { as } n\rightarrow \infty . \end{aligned}$$
(11)
In the second step we apply the Liapounov’s theorem on
$$\begin{aligned} L_i=\frac{1}{n\phi _x(h)}\left( a_3K_i-a_2h^{-1}\beta _iK_i\left( \mathbb {1}_{ \left[ Y_i\le a+b\beta _1\right] }-\alpha \right) \right) . \end{aligned}$$
So, the asymptotic normality, in this lemma, is based on the following result
$$\begin{aligned} \frac{\sum _{i=1}^n\mathbb {E}\left[ |L_i-\mathbb {E}\left[ L_i\right] |^{2+\delta } \right] }{\left( \ var\left( \sum _{i=1}^nL_i\right) \right) ^{(2+\delta )/2}} \rightarrow 0 \hbox { as } n\rightarrow \infty , \quad \hbox {for some } \delta >0. \end{aligned}$$
(12)
From (11), it is clear that
$$\begin{aligned} n\phi _x(h)\ var \left( \sum _{i=1}^nL_i\right) \rightarrow \left[ f^x(t_\alpha (x))(a_1a_3-a_2^2)\sigma (x)\right] ^2, \hbox { as } n \rightarrow \infty . \end{aligned}$$
Therefore, to conclude the proof, it is enough to show that, after normalization, the numerator in (12) converges to 0. Since the observations are independent and identically distributed, by using the \(C_r\)-inequality (see Loève 1963, p. 155), we get
$$\begin{aligned}&(n\phi (h))^{(2+\delta )/2} \sum _{i=1}^n \mathbb {E}\left[ \Big |L_i-\mathbb {E}\left[ L_i\right] \Big |^{2+\delta }\right] \nonumber \\&\quad \le 2^{1+\delta }n\left( n\phi (h)\right) ^{(2+\delta )/2} \left\{ \mathbb {E}\left[ |L_1|^{2+\delta }\right] + |\mathbb {E}\left[ L_1\right] |^{2+\delta }\right\} \nonumber \\&\quad =:M_1+M_2. \end{aligned}$$
Hence,
$$\begin{aligned} M_1= & {} 2^{1+\delta }n^{-\delta /2}(\phi _x(h))^{-1-\delta /2}\mathbb {E}\left[ \left( a_3K_1-Ca_2h^{-1}\beta _1K_1\right) ^{2+\delta }\left| \mathbb {1}_ {\left[ Y_1\le a+b\beta _1\right] }-\alpha \right| ^{2+\delta }\right] \\\le & {} 2^{1+\delta }(n \phi _x(h))^{-\delta /2} \left( \mathbb {E}\left[ \left( a_3K_1-a_2h^{-1}\beta _1K_1\right) ^{2+\delta }\right] /\phi _x(h)\right) \rightarrow 0\hbox { as } n\rightarrow \infty . \end{aligned}$$
Similarly
$$\begin{aligned} M_2= & {} 2^{1+\delta }n^{-\delta /2}\left( \phi _x(h)\right) ^{(2+\delta )/2} \phi _x^{-2-\delta }(h)\mathbb {E}^{2+\delta }\left[ \left( a_3K_1-a_2h^{-1}\beta _1K_1\right) \left| \mathbb {1}_{\left[ Y\le a+b\beta _1\right] }-\alpha \right| \right] \\\le & {} 2^{1+\delta }n^{-\delta /2}\left( \phi _x(h)\right) ^{(2+\delta )/2}\rightarrow 0 \hbox { as } n\rightarrow \infty , \end{aligned}$$
which concludes this proof. \(\square \)
