A nonparametric estimator for the conditional tail index of Pareto-type distributions

Abstract

The tail index is an important parameter in the whole of extreme value theory. In this article, we consider the estimation of the tail index in the presence of a random covariate, where the conditional distribution of the variable of interest is of Pareto-type. More precisely, we use a logarithmic function to link the tail index to the nonlinear predictor induced by covariates, which forms the nonparametric tail index regression models. To estimate the unknown function, we develop an estimation procedure via a local likelihood method. Consistency and asymptotic normality of the estimated functions are established. Subsequently, these theoretical results are illustrated through simulated and real datasets.

Appendix

1.1 Lemmas and their proofs

The following Lemmas will be used in the proofs of Theorem 1 and 2.

Lemma 1

Under Conditions (C.3), (C.6), (C.7) and (C.8), we have

$$\begin{aligned} {\mathbb {E}}\big \{\exp (m(x))\log (Y/u_n)\big |Y>u_n,X=x \big \}= 1+o(\rho (u_{n};x)) \end{aligned}$$

(5.1)

and

$$\begin{aligned} {\mathbb {E}}\big \{[1-\exp (m(x))\log (Y/u_n)]^{2}\big |Y>u_n,X=x\big \}= 1+o(\rho (u_{n};x)) \end{aligned}$$

(5.2)

for any interior point x of $\mathcal {X}$, as $n\rightarrow \infty$.

Proof

By Condition (C.7) and (2.1), it can be shown that

$$\begin{aligned} G_{n}(y;x)&={\mathbb {P}}\big (\exp (m(x))\log (Y/u_n)\le y\big |Y>u_n,X=x\big ) \\&={\mathbb {P}}\big (Y\le u_{n}\exp (y\exp (-m(x)))\big |Y\ge u_{n},X=x\big )\\&=\frac{{\mathbb {P}}\big (u_{n}< Y\le u_{n}\exp (y\exp (-m(x)))\big |X=x\big )}{{\mathbb {P}}\big (Y>u_{n}\big |X=x\big )} \\&=\frac{{\mathbb {P}}\big (Y\le u_{n}\exp (y\exp (-m(x)))\big |X=x\big )-{\mathbb {P}}\big (Y\le u_{n}\big |X=x\big )}{{\mathbb {P}}\big (Y> u_{n}\big |X=x\big )} \\&=\frac{u_{n}^{-\alpha (x)}\ell (u_{n};x)-u_{n}^{-\alpha (x)}\exp (-y)\ell (u_{n}\exp (\exp (-m(x))y);x)}{u_{n}^{-\alpha (x)}\ell (u_{n};x)} \\&=1-\exp (-y)\frac{\ell (u_{n}\exp (\exp (-m(x))y);x)}{\ell (u_{n};x)} \\&=1-\exp (-y)\{1+o(\rho (u_{n};x))\} \end{aligned}$$

for any y, as $n\rightarrow \infty$. Therefore, by Conditions (C.3), (C.6) and (C.8), we have

$$\begin{aligned} {\mathbb {E}}\big \{\exp (m(x))\log (Y/u_n)\big |Y>u_n,X=x \big \}&=\int _{0}^{\infty }(1-G_{n}(y;x))\mathrm {d}y\\&=\int _{0}^{\infty }\exp (-y)\{1+o(\rho (u_{n};x))\}\mathrm {d}y \\&=\int _{0}^{\infty }\exp (-y)\mathrm {d}y + o(\rho (u_{n};x))\\&=1+o(\rho (u_{n};x)). \end{aligned}$$

Similarly, we can also obtain (5.2). $\square$

Lemma 2

Suppose that Conditions (C.5) and (C.8) are satisfied, then

$$\begin{aligned} \frac{n}{n_0}{\mathbb {P}}(Y>u_n|X=x)\rightarrow 1, ~~a.s., ~~\mathrm {as}~ n \rightarrow \infty . \end{aligned}$$

Proof

Under the above Conditions and (2.2) in effect, the proof of the Lemma 2 can immediately be completed from the results that was obtained by Theorem 1 of Wang and Tsai (2009). We here skip the detail. $\square$

1.2 Proofs of the main results

Proof  of  Theorem 3.1

In order to establish the existence and consistency of $\hat{\varvec{\beta }}$, we adapt the proof of Theorem 5.1 in Chapter 6 of Lehmann and Casella (1998). However, before this, we need the following definitions. Define $\varvec{\theta }=S(\varvec{\beta }-\varvec{\beta }^{0})$,  $\hat{\varvec{\theta }}=S(\hat{\varvec{\beta }}-\varvec{\beta }^{0})$, and $\mathbf{U}_i=S^{-1}\mathbf{X}_i=\big (1,\frac{X_i-x}{h},\ldots ,(\frac{X_i-x}{h})^p\big )^{T}$. Then, we obtain

$$\begin{aligned} L_n(\varvec{\theta })&=n_{0}^{-1}\sum _{i=1}^n\{\mathbf{U}_{i}^{T}\varvec{\theta }+\mathbf{X}_{i}^{T}\varvec{\beta ^0}\\&\quad -\exp (\mathbf{U}_{i}^{T}\varvec{\theta }+\mathbf{X}_{i}^{T}\varvec{\beta }^{0}) \log (Y_i/u_n)\}I(Y_i>u_n)K_h(X_i-x). \end{aligned}$$

Thus the problem is equivalent to showing that there exists a solution $\hat{\varvec{\theta }}$ to the likelihood equation

$$\begin{aligned} \frac{\partial L_n(\varvec{\theta })}{\partial \varvec{\theta }}=n_{0}^{-1}\sum _{i=1}^n\{1-\exp (\mathbf{U}_{i}^{T}\varvec{\theta }+\mathbf{X}_{i}^{T}\varvec{\beta ^0})\log (Y_i/u_n)\}\mathbf{U}_iI(Y_i>u_n) K_h(X_i-x)=0\nonumber \\ \end{aligned}$$

(5.3)

such that $\hat{\varvec{\theta }}{\mathop {\rightarrow }\limits ^{P}}\mathbf 0 ,~~ \mathrm {as}~ n\rightarrow \infty$.

Denoted by $Q_\epsilon$ the sphere centered at origin with radius $\epsilon$. We only need to show that for any sufficiently small $\epsilon$, as $n\rightarrow \infty$, the probability tends to 1 that

$$\begin{aligned} L_n(\varvec{\theta })<L_n(\mathbf 0 ) \end{aligned}$$

(5.4)

at all points $\varvec{\theta }$ on the surface of $Q_\epsilon$. Therefore, $L_n(\varvec{\theta })$ has a local maximum in the interior of $Q_\epsilon$. Because at a local maximum the likelihood equation (5.3) must be satisfied, it will follow that for any $\epsilon >0$, with probability tending to 1, the likelihood equation (5.3) has a solution $\hat{\varvec{\theta }}(\epsilon )$ within $Q_\epsilon$. So, let $\hat{\varvec{\theta }}$ be the closest root to 0, then we have ${\mathbb {P}}\{\Vert \hat{\varvec{\theta }}\Vert ^{2}\le \epsilon \}\rightarrow 1, ~~\mathrm {as}~ n\rightarrow \infty$. It implies that $S(\hat{\varvec{\beta }}-\varvec{\beta }^{0}){\mathop {\rightarrow }\limits ^{P}}\mathbf 0 , ~~\mathrm {as}~ n\rightarrow \infty$.

Now we show under assumptions that (5.4) holds. Using Taylor expansion of $L_n(\varvec{\theta })$ around 0, we have that

$$\begin{aligned} L_n(\varvec{\theta })-L_n(\mathbf 0 )=L'_n(\mathbf 0 )^{T}\varvec{\theta }+\frac{1}{2}\varvec{\theta }^{T}L''_n(\mathbf 0 )\varvec{\theta } +R_n(\varvec{\theta }^{*}), \end{aligned}$$

(5.5)

where $\varvec{\theta }^{*}$ lying between $\mathbf 0$ and $\varvec{\theta }$,

$$\begin{aligned} L'_n(\mathbf 0 )&=n_0^{-1}\sum _{i=1}^n\{1-\exp (\mathbf{X}_{i}^{T}\varvec{\beta }^0)\log (Y_i/u_n)\}\mathbf{U}_iI(Y_i>u_n) K_h(X_i-x)\\&=: \left( \frac{n}{n_{0}}\right) \frac{1}{n}Z_{n1},\\ L''_n(\mathbf 0 )&=-n_{0}^{-1}\sum _{i=1}^n \exp (\mathbf{X}_{i}^{T}\varvec{\beta }^0)\log (Y_i/u_n)\mathbf{U}_i\mathbf{U}_i^{T}I(Y_i>u_n) K_h(X_i-x)\\&=: -\left( \frac{n}{n_{0}}\right) \frac{1}{n}Z_{n2}, \end{aligned}$$

and

$$\begin{aligned} R_n(\varvec{\theta }^{*})&=\frac{1}{6}\sum _{j,k,l}\theta _j \theta _k \theta _l\frac{\partial ^{3}L_n(\varvec{\theta }^{*})}{\partial \theta _j \partial \theta _k \partial \theta _l}. \end{aligned}$$

Firstly, by the Strong Law of Large Numbers, we get

$$\begin{aligned} \frac{1}{n}Z_{n1} -{\mathbb {E}}\big [1-\exp (\mathbf{X}_{1}^{T}\varvec{\beta }^0)\log (Y_1/u_n)\big ] \mathbf{U}_1 I(Y_1>u_n)K_h(X_1-x){\mathop {\rightarrow }\limits ^{P}}0, ~~\mathrm {as}~ n\rightarrow \infty . \end{aligned}$$

(5.6)

Next, we have

$$\begin{aligned}&{\mathbb {E}}{\big [1-\exp (\mathbf{X}_{1}^{T}\varvec{\beta }^0)\log (Y_1/u_n)\big ]\mathbf{U}_1 I(Y_1>u_n)K_h(X_1-x)} \\&\quad =\int _{\mathbb {R}}\int _{\mathcal {X}}\big [1-\exp (m(x)+\cdots \\&\qquad +\frac{m^{(p)}(x)}{p!}(t-x)^{p})\log (y/u_n)\big ]\varvec{v}\frac{1}{h} K\left( \frac{t-x}{h}\right) f(y|t)\psi (t) I(y>u_n)\mathrm {d}t\mathrm {d}y \end{aligned}$$

In the last step, we just use a change of variables. By Taylor series expansion and Condition (C.8), the above integral equals

$$\begin{aligned}&\int _{\mathbb {R}}\int _{\mathbb {R}}\left[ 1-\exp (m(x)+\cdots +\frac{m^{(p)}(x)}{p!}(hu)^{p})\log (y/u_n)\right] \\&\qquad \varvec{u}K(u)f(y|x+uh)[\psi (x)+\psi '(x^{*})(hu)] \times \, I(y>u_n)\mathrm {d}u\mathrm {d}y \\&\quad = \int _{\mathbb {R}}\int _{\mathbb {R}}\big [1-\exp (m(x))\log (y/u_n)\big ] \varvec{u}K(u)f(y|x)\psi (x)I(y>u_n)\mathrm {d}u\mathrm {d}y \{1+o(1)\}, \end{aligned}$$

where $x^{*}$ lying between x and $x+uh$. By Condition (C.2) together with (5.6) and Lemma 2, we obtain that

$$\begin{aligned} L'_n(\mathbf 0 ) =&\psi (x){\mathbb {P}}^{-1}(Y>u_n|X=x)\int _{\mathbb {R}}\big [1-\exp (m(x))\log (y/u_n)\big ]f(y|x)I(y>u_n)\mathrm {d}y \\&\times \int _{\mathbb {R}}\varvec{u}K(u)\mathrm {d}u \{1+o_p(1)\}\{1+o(1)\}\\ =\,&\psi (x){\mathbb {P}}^{-1}(Y>u_n|X=x){\mathbb {E}}\big \{\big [1-\exp (m(X))\log (Y/u_n)\big ] I(Y>u_n)\big |X=x \big \}\\&\times \int _{\mathbb {R}}\varvec{u}K(u)\mathrm {d}u \{1+o_p(1)\}\{1+o(1)\}. \end{aligned}$$

Again, by Lemma 1, Conditions (C.1), (C.2) and (C.4), the foregoing expression can be expressed as

$$\begin{aligned}&\psi (x){\mathbb {E}}\{1-\exp (m(X))\log (Y/u_n)|Y>u_n,X=x\}\\&\quad \int _{\mathbb {R}}\varvec{u}K(u)\mathrm {d}u + o_p(\rho (u_{n};x)){\mathop {\rightarrow }\limits ^{P}}0, ~\mathrm {as} \quad n\rightarrow \infty . \end{aligned}$$

Therefore, with probability tending to 1,

$$\begin{aligned} \big |L'_n(\mathbf 0 )^{T}\varvec{\theta }\big |\le \epsilon ^3, ~~\mathrm {as}~ n\rightarrow \infty . \end{aligned}$$

(5.7)

Secondly, similar to (5.6), we can get

$$\begin{aligned} \frac{1}{n}Z_{n2} -{\mathbb {E}}\{ \exp (\mathbf{X}_{1}^{T}\varvec{\beta }^0)\log (Y_1/u_n)\mathbf{U}_1\mathbf{U}_1^{T}I(Y_1>u_n)K_h(X_1-x)\}{\mathop {\rightarrow }\limits ^{P}}0, \end{aligned}$$

(5.8)

where

$$\begin{aligned}&{\mathbb {E}}\{ \exp (\mathbf{X}_{1}^{T}\varvec{\beta }^0)\log (Y_1/u_n)\mathbf{U}_1\mathbf{U}_1^{T}I(Y_1>u_n)K_h(X_1-x)\} \nonumber \\&\quad = \int _{\mathbb {R}}\int _{\mathbb {R}} \exp \big (m(x)+\cdots \nonumber \\&\qquad +\frac{m^{(p)}(x)}{p!}(uh)^{p}\big ) \log (y/u_n)\varvec{u}\varvec{u}^{T}K(u)f(y|x+uh) \psi (x+uh) I(y>u_n)\mathrm {d}u\mathrm {d}y \nonumber \\&\quad = \int _{\mathbb {R}}\int _{\mathbb {R}}\exp (m(x))\log (y/u_n)\varvec{u}\varvec{u}^{T}K(u)f(y|x)\psi (x)I(y>u_n)\mathrm {d}u\mathrm {d}y\{1+o(1)\} \nonumber \\&\quad = \psi (x){\mathbb {E}}\big \{\exp (m(X))\log (Y/u_n)I(Y>u_n)\big |X=x\big \}\Lambda _{0}\{1+o(1)\} \end{aligned}$$

(5.9)

by Conditions (C.2), (C.4) and (C.6). Note that $\Lambda _{0}$ is defined in (3.1). Again, from Lemmas 1 and 2, it follows that

$$\begin{aligned} L''_n(\mathbf 0 )&=-\psi (x){\mathbb {P}}^{-1}(Y>u_n|X=x){\mathbb {E}}\{\exp (m(X))\nonumber \\&\qquad \log (Y/u_n)I(Y>u_n)|X=x\}\Lambda _{0} \{1+o_p(1)\}\{1+o(1)\}\nonumber \\&=-\psi (x){\mathbb {E}}\{\exp (m(X))\log (Y/u_n)|Y>u_n,X=x\}\Lambda _{0}\{1+o_p(1)\}\nonumber \\&=-\psi (x)\Lambda _{0}+o_p(1). \end{aligned}$$

(5.10)

Hence for all $\varvec{\theta }\in Q_\epsilon$,

$$\begin{aligned} \varvec{\theta }^{T}L''_n(\mathbf 0 )\varvec{\theta }< -\lambda \psi (x)\epsilon ^2 \end{aligned}$$

(5.11)

with probability tending to 1, where $\lambda$ is the smallest eigenvalue of $\Lambda _{0}$.

Finally, using the same technical as was used in the proof of $L'_n(\mathbf 0 )$, together with Conditions (C.2), (C.6), (C.8), and Lemma 2, we can show that

$$\begin{aligned} |R_n(\varvec{\theta }^{*})|\le \,&C\epsilon ^{3}\left( \frac{n}{n_0}\right) \frac{1}{n}\sum _{i=1}^n \log (Y_i/u_n)K_h(X_i-x)I(Y_i>u_n)\nonumber \\ =\,&C\epsilon ^{3}\big \{\psi (x){\mathbb {E}}\big (\log (Y_1/u_n)|Y_1>u_n,X_1=x\big )+o_p(1)\big \} \end{aligned}$$

(5.12)

for some constant $C>0$. Thus, substituting (5.7), (5.11) and (5.12) into (5.5), with probability tending to 1, we obtain that

$$\begin{aligned} L_n(\varvec{\theta })-L_n(\mathbf 0 )\le 0 \end{aligned}$$

for all $\varvec{\theta }\in Q_\epsilon$, when $\epsilon$ is small enough. Hence the proof of Theorem 3.1 is completed. $\square$

Proof of Theorem 3.2

By Taylor expansion and Condition (C.5), we have

$$\begin{aligned} \mathbf 0 =L'_{n}(\hat{\varvec{\theta }})&=L'_{n}(\mathbf 0 )+L''_{n}(\mathbf 0 ) \hat{\varvec{\theta }}+O_p(\Vert \hat{\varvec{\theta }}\Vert ^{2})\\&=L'_{n}(\mathbf 0 )+\{L''_{n}(\mathbf 0 )+o_p(1)\}\hat{\varvec{\theta }}. \end{aligned}$$

From (5.10), we conclude that

$$\begin{aligned} \hat{\varvec{\theta }}=-\{L''_{n}(\mathbf 0 )+o_p(1)\}^{-1}L'_{n}(\mathbf 0 ) =-\{-\psi (x)\Lambda _{0}+o_p(1)\}^{-1}L'_{n}(\mathbf 0 ). \end{aligned}$$

(5.13)

On the other hand, to establish asymptotic normality, it remains to calculate the mean and variance of $L'_{n}(0)$, and verify the Lyapounov condition.

By using a Taylor expansion, we get that

$$\begin{aligned} \exp \{m(X_i)\}-\exp \{\mathbf{X}_i^{T}\varvec{\beta }^0\}=\exp (m(x))\frac{m^{(p+1)}(x)}{(p+1)!}(X_i-x)^{p+1}\{1+o_p(1)\}.\nonumber \\ \end{aligned}$$

(5.14)

Next, we evaluate ${\mathbb {E}}{L'_{n}(0)}$, which is given by

$$\begin{aligned} {\mathbb {E}}{L'_{n}(0)} =(nn_0^{-1}){\mathbb {E}}\{1-\exp (\mathbf{X}_1^{T}\varvec{\beta }^{0})\log (Y_1/u_n)\}{} \mathbf{U}_1 I(Y_1>u_n)K_h(X_1-x). \end{aligned}$$

The right-hand side of the above expression can be written as

$$\begin{aligned}&\frac{n}{n_0}{\mathbb {E}}\big \{1-\exp (m(X_1))\log (Y_1/u_n)+\exp (m(X_1))\log (Y_1/u_n)\nonumber \\&\qquad -\,\exp (\mathbf{X}_1^{T}\varvec{\beta }^{0})\log (Y_1/u_n)\big \} \mathbf{U}_1 I(Y_1>u_n)K_h(X_1-x)\nonumber \\&\quad = (nn_0^{-1}){\mathbb {E}}\big \{1-\exp (m(X_1))\log (Y_1/u_n)\big \}{} \mathbf{U}_1 I(Y_1>u_n)K_h(X_1-x)\nonumber \\&\qquad +(nn_0^{-1}){\mathbb {E}}\big \{\exp (m(X_1))\log (Y_1/u_n)\nonumber \\&\quad \quad -\exp (\mathbf{X}_1^{T}\varvec{\beta }^{0})\log (Y_1/u_n)\big \} \mathbf{U}_1 I(Y_1>u_n)K_h(X_1-x) \nonumber \\&\quad =: E_{n1} +E_{n2}. \end{aligned}$$

(5.15)

For $E_{n1}$, similar to the proof of Theorem 1, we have

$$\begin{aligned} E_{n1}&= (nn_0^{-1}){\mathbb {E}}\{1-\exp (m(X_1))\log (Y_1/u_n)\}{} \mathbf{U}_1 I(Y_1>u_n)K_h(X_1-x)\\&= (nn_0^{-1})\int _{\mathbb {R}}\int _{\mathbb {R}}(1-\exp (m(x+uh))\log (y/u_n))\varvec{u}K(u)f(y|x+uh)\\&\quad \psi (x+uh) I(y>u_n)\mathrm {d}u\mathrm {d}y . \end{aligned}$$

Again, by using Taylor’ expansion and noting Conditions (C.2), (C.6), (C.8) and Lemma 2, $E_{n1}$ can be written as

$$\begin{aligned}&(nn_0^{-1})\int _{\mathbb {R}}\int _{\mathbb {R}}(1-\exp \{m(x)+m'(x^{*})(uh)\}\log (y/u_n))\varvec{u}K(u) f(y|x+uh)[\psi (x)\\&\qquad +\,\psi '(x^{*})(uh)] I(y>u_n)\mathrm {d}u\mathrm {d}y\\&\quad = \psi (x){\mathbb {P}}^{-1}(Y>u_n|X=x){\mathbb {E}}\{[1-\exp (m(X))\log (Y/u_n)]I(Y>u_n)\big |X=x\}\\&\qquad \int _{\mathbb {R}}\varvec{u}K(u)\mathrm {d}u\{1+o(1)\}\\&\quad =\psi (x){\mathbb {E}}\{1-\exp (m(X))\log (Y/u_n)\big |Y>u_n,X=x\}\int _{\mathbb {R}}\varvec{u}K(u)\mathrm {d}u\{1+o(1)\}, \end{aligned}$$

where $x^{*}$ is lying between x and $x+uh$. It follows from Lemma 1 that

$$\begin{aligned} E_{n1}= & {} \psi (x) {\mathbb {E}}\{1-\exp (m(X))\log (Y/u_n)|Y>u_n,X=x\}\int _{\mathbb {R}}\varvec{u}K(u)\mathrm {d}u\{1+o(1)\}\nonumber \\= & {} o(\rho (u_{n};x)). \end{aligned}$$

(5.16)

For $E_{n2}$, Conditions (C.2), (C.3), (C.6), (C.8), Lemma 2 and (5.14) implies that

$$\begin{aligned} E_{n2}&= (nn_0^{-1}){\mathbb {E}}\{\exp (m(X_1))\log (Y_1/u_n)\\&\quad -\,\exp (\mathbf{X}_1^{T}\varvec{\beta }^{0})\log (Y_1/u_n)\}{} \mathbf{U}_1 I(Y_1>u_n)K_h(X_1-x)\\&= (nn_0^{-1}){\mathbb {E}}\{ \exp (m(x))\frac{m^{(p+1)}(x)}{(p+1)!}(X_1-x)^{p+1}\log (Y_1/u_n)\\&\quad \mathbf{U}_1 I(Y_1>u_n)K_h(X_1-x)(1+o_p(1))\}\\&= \exp (m(x))\frac{m^{(p+1)}(x)}{(p+1)!}h^{p+1}{\mathbb {P}}^{-1}(Y>u_n|X=x)\int _{\mathbb {R}}\int _{\mathbb {R}} u^{p+1}\varvec{u}K(u)\\&\qquad \log (y/u_n)f(y|x)\psi (x)I(y>u_n)\mathrm {d}u\mathrm {d}y \{1+o(1)\}\\&=\psi (x)\frac{m^{(p+1)}(x)}{(p+1)!}h^{p+1}{\mathbb {P}}^{-1}(Y>u_n|X=x){\mathbb {E}}\{\exp (m(X))\\&\quad \log (Y/u_n)I(Y>u_n)|X=x\}\int _{\mathbb {R}} u^{p+1}\varvec{u}K(u)\mathrm {d}u \{1+o(1)\}. \end{aligned}$$

Then under Lemma 1, we have the following result

$$\begin{aligned} E_{n2}&= \psi (x)\frac{m^{(p+1)}(x)}{(p+1)!}h^{p+1} {\mathbb {E}}\{\exp (m(X))\log (Y/u_n)|Y>u_n,X=x\}\nonumber \\&\quad \int _{\mathbb {R}} u^{p+1}\varvec{u}K(u)\mathrm {d}u\{1+o(1)\}\nonumber \\&= \psi (x)\frac{m^{(p+1)}(x)}{(p+1)!}h^{p+1}\int _{\mathbb {R}} u^{p+1}\varvec{u}K(u)\mathrm {d}u\{1+o(1)\}. \end{aligned}$$

(5.17)

Putting (5.15) together with (5.16) and (5.17) yields

$$\begin{aligned} {\mathbb {E}}{L'_{n}(0)}=\psi (x)\frac{m^{(p+1)}(x)}{(p+1)!}h^{p+1}\int _{\mathbb {R}} u^{p+1}\varvec{u}K(u)\mathrm {d}u+o(h^{p+1})+o(\rho (u_{n};x)). \end{aligned}$$

For the covariance term of $L'_{n}(0)$,

$$\begin{aligned} {\textsf {Cov}}\{L'_{n}(0)\}=\frac{n}{n_0^{2}}\varvec{\triangle }_n-\frac{n}{n_0^{2}}\varvec{r}_n\varvec{r}_n^{T}, \end{aligned}$$

(5.18)

where

$$\begin{aligned} \varvec{\triangle }_n={\mathbb {E}}\{1-\exp (X_1^{T}\beta ^{0})\log (Y_1/u_n)\}^{2}\mathbf{U}_1\mathbf{U}_1^{T} I(Y_1>u_n)K_h^{2}(X_1-x), \end{aligned}$$

and

$$\begin{aligned} \varvec{r}_n={\mathbb {E}}\{1-\exp (\mathbf{X}_1^{T}\varvec{\beta }^{0})\log (Y_1/u_n)\}\mathbf{U}_1 I(Y_1>u_n)K_h(X_1-x). \end{aligned}$$

By the result of ${\mathbb {E}}L'_{n}(0)$, we have that $\frac{n}{n_0}\varvec{r}_n =\psi (x)\frac{m^{(p+1)}(x)}{(p+1)!}h^{p+1}\int u^{p+1}\varvec{u}K(u)du+o(1)$. It means that $\frac{n}{n_0^{2}}\varvec{r}_n\varvec{r}_n^{T}\rightarrow 0$, as $n\rightarrow \infty$.

In addition, we also can show that

$$\begin{aligned} n_0^{-1}\varvec{\triangle }_n =&n_0^{-1} {\mathbb {E}}\{1-\exp (\mathbf{X}_1^{T}\varvec{\beta }^{0})\log (Y_1/u_n)\}^{2}{} \mathbf{U}_1\mathbf{U}_1^{T} I(Y_1>u_n)K_h^{2}(X_1-x) \nonumber \\ =\,&n_0^{-1}\int _{\mathbb {R}}\int _{\mathcal {X}}[1-\exp (m(x)+\cdots +\frac{m^{(p)}(x)}{p!}(t-x)^{p})\log (y/u_n)]^{2}\nonumber \\&\varvec{v}\varvec{v}^{T} \frac{1}{h^2}K^{2}\left( \frac{t-x}{h}\right) f(y|t)\psi (t)I(y>u_n)\mathrm {d}t\mathrm {d}y \nonumber \\ =\,&(n_0h)^{-1} \int _{\mathbb {R}}\int _{\mathbb {R}}[1-\exp (m(x))\log (y/u_n)]^{2}\nonumber \\&\varvec{u}\varvec{u}^{T}K^{2}(u)f(y|x)\psi (x) I(y>u_n)\mathrm {d}u\mathrm {d}y\{1+o(1)\} \nonumber \\ =\,&(n_0h)^{-1}\psi (x){\mathbb {E}}\big \{[1-\exp (m(X))\log (Y/u_n)]^{2}I(Y>u_n)| X=x\big \} \nonumber \\&\Lambda _1 \{1+o(1)\}, \end{aligned}$$

(5.19)

where $\Lambda _1$ is defined in (3.3). So it follows from (5.18), (5.19), Lemmas 1 and 2 that

$$\begin{aligned} {\textsf {Cov}}\{L'_{n}(0)\} =\,&(n_0h)^{-1}\psi (x){\mathbb {P}}^{-1}(Y>u_n|X=x){\mathbb {E}}\{[1-\exp (m(X))\log (Y/u_n)]^{2}\nonumber \\&I(Y>u_n)|X=x\}\Lambda _1 \{1+o(1)\}+o(1)\nonumber \\ =\,&(n_0h)^{-1}\psi (x){\mathbb {E}}\big \{[1-\exp (m(X))\log (Y/u_n)]^{2}|Y>u_n,\nonumber \\ {}&X=x\big \} \Lambda _1\{1+o(1)\}+o(1)\nonumber \\ =\,&(n_0h)^{-1}\psi (x) \Lambda _1+o((n_0h)^{-1}). \end{aligned}$$

(5.20)

On the other hand, to establish the asymptotic normality for $L'_{n}(0)$, it is necessary to show that for any constant vector $\varvec{d}\ne 0$, as $n\rightarrow \infty$ ,

$$\begin{aligned} \sqrt{n_0h}~ \{\varvec{d}^{T}L'_{n}(0)-\varvec{d}^{T}{\mathbb {E}}L'_{n}(0)\} {\mathop {\rightarrow }\limits ^{D}}N(0,\psi (x)\varvec{d}^{T}\Lambda _1\varvec{d}). \end{aligned}$$

(5.21)

The left-hand side of (5.21) can be written as

$$\begin{aligned} \sqrt{n_0h}~n_0^{-1}\sum _{i=1}^{n}\big \{K_h(X_i-x)W_iI(Y_i>u_n)-{\mathbb {E}}K_h(X_i-x)W_iI(Y_i>u_n)\big \}, \end{aligned}$$

where $W_{i}=\big (1-\exp (\mathbf{X}_{i}^{T}\varvec{\beta }^0)\log (Y_i/u_n)\big ){\varvec{d}}^{T}{} \mathbf{U}_i$. Next, it is sufficient to verify the Lyapounov condition:

$$\begin{aligned}&(n_0^{-1}h)^{\frac{3}{2}}n~{\mathbb {E}}\big |K_h(X_i-x)W_iI(Y_i>u_n)\nonumber \\&\quad -{\mathbb {E}}K_h(X_i-x)W_iI(Y_i>u_n)\big |^{3}\rightarrow 0, ~~\mathrm {as}~ n\rightarrow \infty . \end{aligned}$$

(5.22)

Under Conditions (C.2) and (C.6), the left-hand side of (5.22) is bounded by

$$\begin{aligned} 8(n_0^{-1}h)^{\frac{3}{2}}n~{\mathbb {E}}\big |K_h(X_{1}-x)W_{1}\big |^{3}I(Y_{1}>u_n)=O((n_0h)^{-1/2})\rightarrow 0, ~~\mathrm {as}~ n\rightarrow \infty , \end{aligned}$$

and thus the Lyapounov condition hold for the (5.21). By using (5.13) and the slutsky theorem, we have

$$\begin{aligned} \sqrt{n_0h}\{S(\hat{\varvec{\beta }}-\varvec{\beta }^{0}) - \mathscr {B}(x)\}{\mathop {\rightarrow }\limits ^{D}}N\bigg (0, \frac{1}{\psi (x)}\Lambda _0^{-1}\Lambda _1 \Lambda _0^{-1}\bigg ), ~~\mathrm {as}~ n\rightarrow \infty , \end{aligned}$$

where $\mathscr {B}(x)$ is defined in (3.6). We complete the proof of Theorem 3.2. $\square$