Robust conditional Weibull-type estimation

Abstract

We study nonparametric robust tail coefficient estimation when the variable of interest, assumed to be of Weibull type, is observed simultaneously with a random covariate. In particular, we introduce a robust estimator for the tail coefficient, using the idea of the density power divergence, based on the relative excesses above a high threshold. The main asymptotic properties of our estimator are established under very general assumptions. The finite sample performance of the proposed procedure is evaluated by a small simulation experiment.

Appendix

1.1 Proof of Lemma 1

The case $\alpha =\beta =r=0$ is trivial, so we only consider case (ii). Let $p_n:=F(u_n;x)$. We remark that

$$\begin{aligned} m(u_n,\alpha , \beta , r;x)&= {\mathbb {E}}\left( \hbox {e}^{-c_n\alpha \left[ \left( {Q(U;x) \over Q(p_n;x)}\right) ^{1/\theta (x)}-1\right] } \left( {Q(U;x) \over Q(p_n;x)}\right) ^{\beta } \right. \\&\left. \times \left( \ln {Q(U;x) \over Q(p_n;x)}\right) ^r_+ 1\!\!1_{\{Q(U;x)>Q(p_n;x)\}}\right) \\&= \int _{p_n}^{\widetilde{p}_n} \hbox {e}^{-c_n\alpha \left[ \left( {Q(u;x) \over Q(p_n;x)}\right) ^{1/\theta (x)}-1\right] } \left( {Q(u;x) \over Q(p_n;x)}\right) ^{\beta }\left( \ln {Q(u;x) \over Q(p_n;x)}\right) ^r~\hbox {d}u\\&{+}\int _{\widetilde{p}_n}^1 \hbox {e}^{-c_n\alpha \left[ \left( {Q(u;x) \over Q(p_n;x)}\right) ^{1/\theta (x)}-1\right] } \left( {Q(u;x)\!\over Q(p_n;x)}\right) ^{\beta }\!\left( \ln {Q(u;x) \over Q(p_n;x)}\right) ^r \hbox {d}u\\&=: m^{(1)}(u_n, \alpha ,\beta , r;x) + m^{(2)}(u_n,\alpha , \beta , r;x), \end{aligned}$$

where $U$ is a uniform $[0,1]$ random variable and $\widetilde{p}_n := 1-{1-p_n \over \ln {e \over 1-p_n}}$.

We will study the two terms separately. First, we remark that

$$\begin{aligned} {Q(u;x) \over Q(p_n;x)} = \left( 1+{-\ln {1-u \over 1-p_n} \over -\ln (1-p_n)}\right) ^{\theta (x)} {\ell \left( \left( 1+{-\ln {1-u \over 1-p_n} \over -\ln (1-p_n)}\right) (-\ln (1-p_n));x\right) \over \ell (-\ln (1-p_n);x)}. \nonumber \\ \end{aligned}$$

(9)

Thus by the change of variable $z={1-u \over 1-p_n}$, Assumption $({\mathcal {R}})$ and the bound ${\rho (x) - 1 \over 2} z^2 \le D_{\rho (x)}(1+z)-z\le 0$, for $z \ge 0$, we deduce that

$$\begin{aligned}&m^{(1)}(u_n, \alpha ,\beta , r;x) =(1-p_n)\\&\qquad \times \,\int _{{1-\widetilde{p}_n \over 1-p_n}}^{1} \hbox {e}^{-c_n\alpha \left[ \left( 1+{-\ln {z} \over c_n}\right) \left( 1+b(c_n;x)D_{\rho (x)}\left( 1+{-\ln {z} \over c_n}\right) (1+o(1))\right) ^{1/\theta (x)}-1\right] }\\&\qquad \times \left( 1+{-\ln {z} \over c_n}\right) ^{\theta (x) \beta } \left( 1+b(c_n;x)D_{\rho (x)}\left( 1+{-\ln {z} \over c_n}\right) (1+o(1))\right) ^{\beta }\\&\qquad \times \left( \ln \left[ \left( 1{+}{-\ln {z} \over c_n}\right) ^{\theta (x)} \left( 1{+}b(c_n;x)D_{\rho (x)}\left( 1+{-\ln {z} \over c_n}\right) (1+o(1)) \right) \right] \right) ^r~\hbox {d}z\\&\quad =(1-p_n)\int _{{1-\widetilde{p}_n \over 1-p_n}}^{1} z^\alpha \left[ \theta ^r(x) \left( {-\ln z \over c_n}\right) ^r + \theta ^{r}(x)\left( \theta (x)\beta -{r\over 2}\right) \left( {-\ln z\over c_n}\right) ^{r+1}\right. \\&\qquad +\,\, r \theta ^{r-1}(x) \left( {-\ln z \over c_n}\right) ^r b(c_n;x)(1+o(1))- \alpha \theta ^{r-1}(x) {(-\ln z)^{r+1}\over c^r_n} b(c_n;x)\\&\left. \qquad \times \,(1+o(1))+O\left( \left( {-\ln z \over c_n}\right) ^{r+2}\right) \right] ~\hbox {d}z. \end{aligned}$$

Now, we remark that

$$\begin{aligned} \int _{{1-\widetilde{p}_n \over 1-p_n}}^{1} z^\alpha (-\ln z)^r\hbox {d}z = {1 \over (1+\alpha )^{r+1}}\left\{ \Gamma (r+1)- \Gamma (r+1,(1+\alpha )\ln (1+c_n)) \right\} . \end{aligned}$$

Thus,

$$\begin{aligned}&m^{(1)}(u_n,\alpha ,\beta ,r;x)=(1-p_n) {\Gamma (1+r) \over (1+\alpha )^{1+r}} \theta ^{r}(x)\\&\quad \times \, \left\{ c_n^{-r} + {\theta (x)\beta \over 1+\alpha } \, c_n^{-1} 1\!\!1_{\{r=0\}} +{r-\alpha \over 1+\alpha } {b(c_n;x) \over \theta (x)} c_n^{-r}\right. \\&\quad \left. -c_n^{-1-\alpha }1\!\!1_{\{r=0\}}+o(b(c_n;x)c_n^{-r}) +O\left( {\ln c_n \over c_n^{2+\alpha }}1\!\!1_{\{r=0\}}\right) +O\left( {1\over c_n^{2}} 1\!\!1_{\{r=0\}}\right) \right. \\&\quad \left. +O\left( {(\ln c_n)^r \over c_n^{1+r+\alpha }}1\!\!1_{\{r>0\}}\right) +O\left( {1\over c_n^{1+r}} 1\!\!1_{\{r>0\}}\right) \right\} . \end{aligned}$$

Now, concerning the $m^{(2)}(u_n,\alpha , \beta ,r;x)$ term, using the monotonicity of $Q$ and of the exponential function leads to the inequality

$$\begin{aligned} m^{(2)}(u_n,\alpha ,\beta ,r;x)&\le \hbox {e}^{-c_n \alpha \left[ \left( {Q(\!\widetilde{p}_n;x) \over Q(p_n;x)}\right) ^{1/\theta (x)}-1\right] } \!\int _{\widetilde{p}_n}^1 \left( {Q(u;x) \over Q(p_n;x)}\right) ^{\beta } \left( \!\ln {Q(u;x) \over Q(p_n;x)} \right) ^r\hbox {d}u\\&=: T_1 \times T_2. \end{aligned}$$

Clearly, using (9), Assumption $({\mathcal {R}})$ and the bound for $D_{\rho (x)}(1+.)$, we have

$$\begin{aligned} \left( {Q(\widetilde{p}_n;x) \over Q(p_n;x)} \right) ^{1/\theta (x)} =1+{b\left( c_n;x\right) \over \theta (x)} {\ln (1+c_n) \over c_n}(1+o(1))+{\ln (1+c_n) \over c_n}. \end{aligned}$$

This implies that

$$\begin{aligned} T_1&= \hbox {e}^{-\alpha \ln (1+c_n)} \hbox {e}^{-{\alpha \over \theta (x)} b(c_n;x) \ln (1+c_n)(1+o(1))}= c_n^{-\alpha } (1+o(1)) \end{aligned}$$

since $\rho (x)<0$.

Now, concerning the term $T_2$, using the tail quantile function $U(y; x) := Q\left( 1-{1 \over y}; x\right) ,\,y >1$, combined with the change of variables $z={1-p_n \over 1-u}$, we deduce that

$$\begin{aligned} T_2&= (1-p_n)\left( {a\left( {1 \over 1-p_n};x \right) \over U\left( {1 \over 1-p_n};x \right) } \right) ^r\\&\times \, \int _{1+c_n}^\infty \left[ 1+ {a\left( {1 \over 1-p_n};x \right) \over U\left( {1 \over 1-p_n};x \right) } {U\left( {z \over 1-p_n}; x\right) -U\left( {1 \over 1-p_n};x \right) \over a\left( {1\over 1-p_n}; x\right) }\right] ^{\beta } {1\over z^2} \\&\times \left( {\ln U\left( {z \over 1-p_n}; x\right) -\ln U\left( {1 \over 1-p_n};x \right) \over a\left( {1 \over 1-p_n};x \right) / U\left( {1 \over 1-p_n};x \right) } \right) ^r\hbox {d}z, \end{aligned}$$

where $a$ is the positive function that appears in the max-domain of attraction condition

$$\begin{aligned} \frac{U(tx)-U(t)}{a(t)} \rightarrow \ln x,\quad \hbox {as } t \rightarrow \infty ,\, \hbox {for all } x >0. \end{aligned}$$

We have to study two cases depending on the sign of $\beta$.

First case: $\beta \le 0$. Using the fact that $U(.)$ is an increasing function combined with Corollary B.2.10 in de Haan and Ferreira (2006, p. 376), we deduce that for $p_n$ sufficiently large and $\varepsilon$ sufficiently small that

$$\begin{aligned} T_2\le (1-p_n) \left( {a\left( {1 \over 1-p_n};x \right) \over U\left( {1 \over 1-p_n};x \right) } \right) ^r O\left( c_n^{r\varepsilon -1}\right) = O\left( {1-p_n \over c_n^{1+r-r\varepsilon }}\right) , \end{aligned}$$

where we have also used that

$$\begin{aligned} {a\left( {1 \over 1-p_n};x \right) \over U\left( {1 \over 1-p_n};x \right) } =O\left( \frac{1}{c_n}\right) ,\quad \hbox {as }p_n \uparrow 1, \end{aligned}$$

see, e.g., the proof of Lemma 1 in de Wet et al. (2013).

Second case: $\beta >0$. Using again Corollary B.2.10 in de Haan and Ferreira (2006, p. 376), we have for $p_n$ sufficiently large, $\delta$ and $\widetilde{\delta }$ positive constants, and $\varepsilon$ and $\widetilde{\varepsilon }$ sufficiently small that

$$\begin{aligned} T_2&\le (1-p_n) \delta ^r \, \left( {a\left( {1 \over 1-p_n};x \right) \over U\left( {1 \over 1-p_n};x \right) }\right) ^{r+\beta } \widetilde{\delta }^{\beta }\\&\times \, \left[ 1+ {U\left( {1 \over 1-p_n};x \right) \over a\left( {1 \over 1-p_n};x \right) } {1 \over \widetilde{\delta }(1+c_n)^{\widetilde{\varepsilon }}} \right] ^{\beta } \int _{1+c_n}^\infty z^{\beta \widetilde{\varepsilon }+r\varepsilon -2}~\hbox {d}z\\&= (1-p_n) \delta ^r \, {1\over (1+c_n)^{\widetilde{\varepsilon } \beta }}\left( {a\left( {1 \over 1-p_n};x \right) \over U\left( {1 \over 1-p_n};x \right) }\right) ^{r}\\&\times \, \left[ 1+ {a\left( {1 \over 1-p_n};x \right) \over U\left( {1 \over 1-p_n};x \right) } \widetilde{\delta }(1+c_n)^{\widetilde{\varepsilon }} \right] ^{\beta } \int _{1+c_n}^\infty z^{\beta \widetilde{\varepsilon }+r\varepsilon -2}~\hbox {d}z = O\left( {1-p_n \over c_n^{1+r-r\varepsilon }}\right) . \end{aligned}$$

Finally

$$\begin{aligned} m^{(2)}(u_n,\alpha ,\beta ,r;x) = O\left( {1-p_n \over c_n^{1+r+\alpha -r\varepsilon }}\right) . \end{aligned}$$

Combining all these results together leads to Lemma 1. $\square$

1.2 Proof of Lemma 2

From the rule of repeated expectations, we have that

$$\begin{aligned} m_n(K,\alpha ,\beta ,r;x) = {\mathbb {E}}(K_h(x-X)m(u_n,\alpha ,\beta ,r;X)). \end{aligned}$$

Straightforward operations give

$$\begin{aligned}&m_n(K,\alpha ,\beta ,r;x)=\int _\Omega K(z)m(u_n,\alpha ,\beta ,r;x-hz)f(x-hz)~\hbox {d}z \\&\quad =m(u_n,\alpha ,\beta ,r;x)f(x) +m(u_n,\alpha ,\beta ,r;x) \int _\Omega K(z)(f(x-hz)-f(x))~\hbox {d}z \\&\qquad + f(x)\int _\Omega K(z)(m(u_n,\alpha ,\beta ,r;x-hz)-m(u_n,\alpha ,\beta ,r;x))~\hbox {d}z\\&\qquad + \int _\Omega K(z)(m(u_n,\alpha ,\beta ,r;x-hz) -m(u_n,\alpha ,\beta ,r;x))(f(x-hz)-f(x))~\hbox {d}z\\&\quad =: m(u_n,\alpha ,\beta ,r;x)f(x)+T_3+T_4+T_5. \end{aligned}$$

We now analyze each of the terms separately. By $({\mathcal {F}})$ and $({\mathcal {K}}),$ we have that

$$\begin{aligned} |T_{3}|&\le m(u_n,\alpha ,\beta ,r;x)M_f \int _\Omega K(z) \Vert hz\Vert ^{\eta _f}~\hbox {d}z \\&= O(m(u_n,\alpha ,\beta ,r;x) h^{\eta _f}), \end{aligned}$$

and, by $({\mathcal {M}})$ and $({\mathcal {K}})$,

$$\begin{aligned} |T_4|&\le f(x) m(u_n,\alpha ,\beta ,r;x) \int _\Omega K(z)\left| \frac{m(u_n,\alpha \beta ,r;x-hz)}{m(u_n,\alpha ,\beta ,r;x)}-1\right| ~\hbox {d}z \\&= O(m(u_n,\alpha ,\beta ,r;x)\Phi _n(x)). \end{aligned}$$

Using similar arguments, one obtains $T_5=O(m(u_n,\alpha ,\beta ,r;x)h^{\eta _f}\Phi _n(x))$. This proves the statement about the unconditional expectation.

For what concerns the convergence in probability, we already have from the first part of the proof that

$$\begin{aligned} {\mathbb {E}} \left( \widetilde{T}_n(K,\alpha ,\beta ,r;x) \right) = \frac{\theta ^r(x)\Gamma (1+r)}{(1+\alpha )^{1+r}}\;(1+o(1)). \end{aligned}$$

Also, again by using the result from the first part of the proof

$$\begin{aligned}&{\mathbb {V}} \text{ ar }\left( \widetilde{T}_n(K,\alpha ,\beta ,r;x)\right) \\&\quad = \frac{c_n^{2r}\; {\mathbb {V}}\text{ ar } \left( K_h(x-X) \mathrm{e}^{-c_n\alpha \left[ \left( {Y \over u_n}\right) ^{1/\theta (x)}-1\right] } \left( {Y \over u_n}\right) ^{\beta } \left( \ln {Y \over u_n}\right) ^r_+ 1\!\!1_{\{Y>u_n\}} \right) }{n ({\overline{F}}(u_n;x)f(x))^2} \\&\quad = \frac{\theta ^{2r}(x)\Vert K\Vert _2^2\Gamma (1+2r)}{(1+2\alpha )^{1+2r}nh^p {\overline{F}}(u_n;x)f(x) }\;(1+o(1)). \end{aligned}$$

Thus,

$$\begin{aligned} {\mathbb {V}} \text{ ar }\left( \widetilde{T}_n(K,\alpha ,\beta ,r;x)\right) \rightarrow 0 \end{aligned}$$

under the assumptions of the lemma and the convergence in probability follows. $\square$

1.3 Proof of Corollary 1

First, note that

$$\begin{aligned} {\widehat{f}}_n(x) := \frac{1}{n} \sum _{i=1}^n K_h(x-X_i), \end{aligned}$$

is a classical kernel density estimator for $f$. As shown in Parzen (1962), if $nh^p \rightarrow \infty$, then for all $x \in {\mathbb {R}}^p$ where $f(x)>0$ one has that $\widehat{f}_n(x) \mathop {\rightarrow }\limits ^{{\mathbb {P}}} f(x)$. The result follows then by noting that

$$\begin{aligned} \frac{\widehat{{\overline{F}}}(u_n;x)}{{\overline{F}}(u_n;x)} =\frac{f(x)}{{\widehat{f}}_n(x)}{\widetilde{T}}_n(K,0,0,0;x). \end{aligned}$$

1.4 Proof of Theorem 1

To prove the theorem, we will adjust the arguments used to prove the existence and consistency of solutions of the likelihood estimating equation, see e.g., Theorem 3.7 and Theorem 5.1 in Chapter 6 of Lehmann and Casella (1998), to the MDPD framework. We rescale the objective function ${\widehat{\Delta }}_\alpha (\theta ;{\widehat{c}}_n)$ as

$$\begin{aligned} {\widetilde{\Delta }}_\alpha (\theta ;{\widehat{c}}_n) := \frac{\widehat{\Delta }_\alpha (\theta ;{\widehat{c}}_n)}{{\overline{F}}(u_n;x)f(x)c_n^\alpha }. \end{aligned}$$

First, we will show that

$$\begin{aligned} {\mathbb {P}}_{\theta _0(x)}(\widetilde{\Delta }_\alpha (\theta _0(x); {\widehat{c}}_n) < \widetilde{\Delta }_\alpha (\theta ;{\widehat{c}}_n)) \rightarrow 1 \end{aligned}$$

(10)

as $n \rightarrow \infty$, for any $\theta$ sufficiently close to $\theta _0(x)$.

By Taylor’s theorem

$$\begin{aligned} {\widetilde{\Delta }}_\alpha \left( \theta ;{\widehat{c}}_n\right) - {\widetilde{\Delta }}_\alpha \left( \theta _0(x);{\widehat{c}}_n\right)&= {\widetilde{\Delta }}_\alpha ^\prime \left( \theta _0(x); {\widehat{c}}_n\right) (\theta -\theta _0(x)) +\frac{1}{2} \widetilde{\Delta }_\alpha ^{\prime \prime }\left( \theta _0(x);{\widehat{c}}_n\right) \\&\times \,(\theta -\theta _0(x))^2 +\frac{1}{6} \widetilde{\Delta }_\alpha ^{\prime \prime \prime }\left( \widetilde{\theta };{\widehat{c}}_n\right) (\theta -\theta _0(x))^3, \end{aligned}$$

where ${\widetilde{\theta }}$ is a value between $\theta$ and $\theta _0(x)$. The term ${\widetilde{\Delta }}_\alpha ^\prime (\theta _0(x);{\widehat{c}}_n)$ can be obtained from (4). Write $\widetilde{\Delta }_\alpha ^\prime (\theta _0(x);{\widehat{c}}_n)=: R_1+R_2+R_3-R_4$. For analyzing the term $R_1$, we use the recursive relationships

$$\begin{aligned} \Gamma (a,b)&= \hbox {e}^{-b}b^{a-1}+(a-1)\Gamma (a-1,b), \\ \Psi (a,b)&= \hbox {e}^{-b}b^{a-1}\ln b +(a-1)\Psi (a-1,b)+\Gamma (a-1,b), \end{aligned}$$

Lemma 2, and the consistency of $\widehat{\overline{F}}(u_n;x)$, giving

$$\begin{aligned} R_1 \mathop {\rightarrow }\limits ^{{\mathbb {P}}} -\frac{\alpha }{\theta _0^{\alpha +1}(x)(1+\alpha )}. \end{aligned}$$

For $R_2$ we rearrange the terms to obtain

$$\begin{aligned} R_2&= \frac{1+\alpha }{\theta _0^{\alpha +1}(x)}\, (1+o_{\mathbb {P}}(1)) \Biggr \{\frac{T_n(K,\alpha , \alpha (1/\theta _0(x)-1),0;x)}{{\overline{F}}(u_n;x)f(x)} \\&\left. +\frac{\frac{1}{n} \sum _{i=1}^{n} K_h(x-X_i)\left[ \hbox {e}^{-{\widehat{c}}_n\alpha \left[ \left( {Y_i \over u_n}\right) ^{1/\theta _0(x)}-1\right] }-\hbox {e}^{-c_n\alpha \left[ \left( {Y_i \over u_n}\right) ^{1/\theta _0(x)}-1\right] } \right] \left( \frac{Y_i}{u_n} \right) ^{\alpha (1/\theta _0(x)-1)}1\!\!1_{\lbrace Y_i > u_n \rbrace }}{{\overline{F}}(u_n;x)f(x)} \right\} \\&=: \frac{1+\alpha }{\theta _0^{\alpha +1}(x)} (R_{2,1}+R_{2,2})(1+o_{\mathbb {P}}(1)). \end{aligned}$$

By Lemma 2, we have that $R_{2,1} \mathop {\rightarrow }\limits ^{{\mathbb {P}}} (1+\alpha )^{-1}$. For the term $R_{2,2}$, we use the mean value theorem to obtain, with $\widetilde{c}_n$ being a random value between $c_n$ and ${\widehat{c}}_n$,

$$\begin{aligned} R_{2,2}&= \alpha \ln \frac{\widehat{{\overline{F}}}(u_n;x)}{{\overline{F}}(u_n;x)}\\&{\times } \left[ \frac{\frac{1}{n} \sum _{i=1}^{n} K_h(x-X_i) \hbox {e}^{-\widetilde{c}_n\alpha \left[ \left( {Y_i \over u_n}\right) ^{1/\theta _0(x)}\!-\!1\right] } \left( \frac{Y_i}{u_n} \right) ^{\alpha (1/\theta _0(x)-1)\!+\!1/\theta _0(x)}1\!\!1_{\lbrace Y_i > u_n \rbrace }}{{\overline{F}}(u_n;x)f(x)}\right. \\&-\left. \frac{\frac{1}{n} \sum _{i=1}^{n} K_h(x-X_i) \hbox {e}^{-\widetilde{c}_n\alpha \left[ \left( {Y_i \over u_n}\right) ^{1/\theta _0(x)}-1\right] } \left( \frac{Y_i}{u_n} \right) ^{\alpha (1/\theta _0(x)-1)}1\!\!1_{\lbrace Y_i > u_n \rbrace }}{{\overline{F}}(u_n;x)f(x)} \right] \\&=: \alpha \ln \frac{\widehat{{\overline{F}}}(u_n;x)}{{\overline{F}}(u_n;x)} (R_{2,2,1}-R_{2,2,2}), \end{aligned}$$

which can be easily bounded as follows:

$$\begin{aligned} R_{2,2,1}&\le \frac{\frac{1}{n} \sum _{i=1}^{n} K_h(x-X_i) \left( \frac{Y_i}{u_n}\right) ^{\alpha (1/\theta _0(x)-1)+1/\theta _0(x)}1\!\!1_{\lbrace Y_i>u_n\rbrace }}{{\overline{F}}(u_n;x)f(x)}=O_{\mathbb {P}}(1), \\ R_{2,2,2}&\le \frac{\frac{1}{n} \sum _{i=1}^{n} K_h(x-X_i) \left( \frac{Y_i}{u_n}\right) ^{\alpha (1/\theta _0(x)-1)}1\!\!1_{\lbrace Y_i>u_n \rbrace }}{{\overline{F}}(u_n;x)f(x)}=O_{\mathbb {P}}(1), \end{aligned}$$

and, therefore, by the consistency of $\widehat{{\overline{F}}}(u_n;x)$, the convergence $R_{2,2} \mathop {\rightarrow }\limits ^{{\mathbb {P}}}0$ follows. Combining all results gives

$$\begin{aligned} R_2 \mathop {\rightarrow }\limits ^{{\mathbb {P}}} \frac{1}{\theta _0^{\alpha +1}(x)}. \end{aligned}$$

The terms $R_3$ and $R_4$ can be analyzed in an analogous way and yield

$$\begin{aligned} R_3 \mathop {\rightarrow }\limits ^{{\mathbb {P}}} 0 \quad \hbox {and} \quad R_4 \mathop {\rightarrow }\limits ^{{\mathbb {P}}}\frac{1}{\theta _0^{\alpha +1}(x)(1+\alpha )}. \end{aligned}$$

Thus, $\widetilde{\Delta }_\alpha ^\prime (\theta _0(x);{\widehat{c}}_n) \mathop {\rightarrow }\limits ^{{\mathbb {P}}} 0$. Let $|\theta -\theta _0(x)|=r,\,r>0$. With probability tending to 1, we have that

$$\begin{aligned} \left| \widetilde{\Delta }_\alpha ^\prime (\theta _0(x);{\widehat{c}}_n)(\theta -\theta _0(x)) \right| < r^3. \end{aligned}$$

We now turn to the analysis of $\widetilde{\Delta }_\alpha ^{\prime \prime }(\theta _0(x);{\widehat{c}}_n)$. Let

$$\begin{aligned} \phi (a,b) := \int _b^\infty \ln ^2 z \; z^{a-1} \hbox {e}^{-z}~\hbox {d}z, \end{aligned}$$

and

$$\begin{aligned}&\widehat{T}_n(K,\alpha ,\beta ,r;x) := \frac{1}{n} \sum _{i=1}^{n} K_h(x-X_i)\hbox {e}^{-{\widehat{c}}_n\alpha \left[ \left( {Y_i \over u_n}\right) ^{1/\theta _0(x)}-1\right] }\\&\quad \times \, \left( \frac{Y_i}{u_n} \right) ^{\beta }\left( \ln \frac{Y_i}{u_n}\right) _+^r1\!\!1_{\lbrace Y_i > u_n \rbrace }. \end{aligned}$$

Note that the function $\phi (a,b)$ satisfies the recursive relationship

$$\begin{aligned} \phi (a,b)=\hbox {e}^{-b} b^{a-1}\ln ^2 b+(a-1)\phi (a-1,b)+2\Psi (a-1,b). \end{aligned}$$

(11)

After tedious calculations, one obtains the following expression for $\widetilde{\Delta }_\alpha ^{\prime \prime }(\theta _0(x);{\widehat{c}}_n)$:

$$\begin{aligned}&\widetilde{\Delta }_\alpha ^{\prime \prime }(\theta _0(x);{\widehat{c}}_n)\\&\quad =\frac{T_n(K,0,0,0;x)}{{\overline{F}}(u_n;x)f(x)} \frac{\mathrm{e}^{{\widehat{c}}_n(1+\alpha )}{\widehat{c}}_n^{~\alpha \theta _0(x)}}{\theta _0^{\alpha +2}(x)(1+ \alpha )^{1+\alpha (1-\theta _0(x))}c_n^\alpha }\\&\qquad \times \,\, \lbrace \alpha (1+\alpha )\Gamma (\alpha (1-\theta _0(x))+1, {\widehat{c}}_n(1+\alpha )) +2\alpha ^2 \theta _0(x)\Psi (\alpha (1-\theta _0(x))\\&\qquad +\,\,1,{\widehat{c}}_n(1+\alpha )) +\alpha ^2 \theta _0^2(x) \phi (\alpha (1-\theta _0(x))+1,{\widehat{c}}_n(1+\alpha )) \\&\qquad -\,\,2\alpha ^2 \theta _0(x)\ln ({\widehat{c}}_n(1+\alpha )) [\Gamma (\alpha (1-\theta _0(x))+1,{\widehat{c}}_n(1+\alpha ))\\&\qquad +\,\,\theta _0(x)\Psi (\alpha (1-\theta _0(x))+1,{\widehat{c}}_n(1+\alpha ))] \\&\qquad +\,\,\alpha ^2 \theta _0^2(x) \ln ^2({\widehat{c}}_n(1+\alpha )) \Gamma (\alpha (1-\theta _0(x))+1,{\widehat{c}}_n(1+\alpha ))\rbrace \\&\qquad -\,\,\frac{(\alpha +1)^2{\widehat{c}}_n^{~\alpha }}{\theta _0^{\alpha +2}(x)c_n^\alpha } \frac{\widehat{T}_n(K,\alpha ,\alpha (1/\theta _0(x)-1),0;x)}{{\overline{F}}(u_n;x)f(x)} \\&\qquad -\,\,\frac{2(\alpha +1)^2{\widehat{c}}_n^{~\alpha }}{\theta _0^{\alpha +3}(x)c_n^\alpha } \frac{\widehat{T}_n(K,\alpha ,\alpha (1/\theta _0(x)-1),1;x)}{{\overline{F}}(u_n;x)f(x)} \\&\qquad +\,\,\frac{2(\alpha +1)^2{\widehat{c}}_n^{~\alpha +1}}{\theta _0^{\alpha +3}(x)c_n^\alpha } \frac{\widehat{T}_n(K,\alpha ,\alpha (1/\theta _0(x)-1)+1/\theta _0(x),1;x)}{{\overline{F}}(u_n;x)f(x)} \\&\qquad -\,\,\frac{\alpha (1+\alpha ){\widehat{c}}_n^{~\alpha }}{\theta _0^{\alpha +4}(x)c_n^\alpha } \frac{\widehat{T}_n(K,\alpha ,\alpha (1/\theta _0(x)-1),2;x)}{{\overline{F}}(u_n;x)f(x)} \\&\qquad +\,\,\frac{(1+2\alpha )(1+\alpha ){\widehat{c}}_n^{~\alpha +1}}{\theta _0^{\alpha +4}(x)c_n^\alpha } \frac{\widehat{T}_n(K,\alpha ,\alpha (1/\theta _0(x)-1)+1/\theta _0(x),2;x)}{{\overline{F}}(u_n;x)f(x)} \\&\qquad -\,\,\frac{\alpha (1+\alpha ){\widehat{c}}_n^{~\alpha +2}}{\theta _0^{\alpha +4}(x)c_n^\alpha } \frac{\widehat{T}_n(K,\alpha ,\alpha (1/\theta _0(x)-1)+2/\theta _0(x),2;x)}{{\overline{F}}(u_n;x)f(x)}. \end{aligned}$$

By a line of argumentation similar to that used for $\widetilde{\Delta }_\alpha ^\prime (\theta _0(x);{\widehat{c}}_n)$ and also using (11), one obtains that under the conditions of the theorem

$$\begin{aligned} \widetilde{\Delta }_\alpha ^{\prime \prime }(\theta _0(x);{\widehat{c}}_n) \mathop {\rightarrow }\limits ^{{\mathbb {P}}} \frac{1+\alpha ^2}{\theta _0^{\alpha +2}(x)(1+\alpha )^2}. \end{aligned}$$

(12)

Write

$$\begin{aligned}&\frac{1}{2} \widetilde{\Delta }_\alpha ^{\prime \prime }(\theta _0(x);{\widehat{c}}_n)(\theta -\theta _0(x))^2 = \frac{1+\alpha ^2}{2\theta _0^{\alpha +2}(x)(1+\alpha )^2}(\theta -\theta _0(x))^2 \\&\quad +\,\,\frac{1}{2} \left( \widetilde{\Delta }_\alpha ^{\prime \prime }(\theta _0(x);{\widehat{c}}_n) - \frac{1+\alpha ^2}{\theta _0^{\alpha +2}(x)(1+\alpha )^2} \right) (\theta -\theta _0(x))^2. \end{aligned}$$

The random part in the right-hand side of the above display is in absolute value less than $r^3$ with probability tending to 1. There exist thus a $\delta _1 >0$ and an $r_0>0$ such that for $r<r_0$

$$\begin{aligned} \frac{1}{2} \widetilde{\Delta }_\alpha ^{\prime \prime }(\theta _0(x);{\widehat{c}}_n)(\theta -\theta _0(x))^2 > \delta _1 r^2 \end{aligned}$$

with probability tending to 1.

For the third-order derivative, one can show that $|\widetilde{\Delta }_\alpha ^{\prime \prime \prime }(\theta ;{\widehat{c}}_n)| \le M(\varvec{V})$, where $\varvec{V} :=[(X_1,Y_1),\ldots , (X_n,Y_n)]$, for $\theta \in (\theta _0(x)-r,\theta _0(x)+r)$, with $M(\varvec{V}) \mathop {\rightarrow }\limits ^{{\mathbb {P}}} M$, which is bounded. The derivation is straightforward but lengthy and is therefore omitted from the paper. We can thus conclude that with probability tending to 1,

$$\begin{aligned} \frac{1}{6}|\widetilde{\Delta }_\alpha ^{\prime \prime \prime }(\widetilde{\theta };{\widehat{c}}_n)(\theta -\theta _0(x))^3| < \frac{1}{3}\; M r^3. \end{aligned}$$

Overall, we have that with probability tending to 1,

$$\begin{aligned} \widetilde{\Delta }_\alpha (\theta ;{\widehat{c}}_n)-\widetilde{\Delta }_\alpha (\theta _0(x);{\widehat{c}}_n) > \delta _1r^2-(1+M/3)r^3, \end{aligned}$$

which is positive if $r < \delta _1/(1+M/3)$ and thus (10) follows.

To complete the proof, we adjust the line of argumentation of Theorem 3.7 in Chapter 6 of Lehmann and Casella (1998). Let $\delta >0$ be such that $\theta _0(x)-\delta >0$ and define

$$\begin{aligned}&S_n(\delta ) = \left\{ {\varvec{v}} : \widetilde{\Delta }_\alpha (\theta _0(x);{\widehat{c}}_n) < \widetilde{\Delta }_\alpha (\theta _0(x)-\delta ;{\widehat{c}}_n)\right. \\&\left. \hbox { and } \quad \widetilde{\Delta }_\alpha (\theta _0(x);{\widehat{c}}_n) < \widetilde{\Delta }_\alpha (\theta _0(x)+\delta ;{\widehat{c}}_n)\right\} . \end{aligned}$$

For any ${\varvec{v}} \in S_n(\delta )$, since $\widetilde{\Delta }_\alpha (\theta ; {\widehat{c}}_n)$ is differentiable with respect to $\theta$, we have that there exists a $\widehat{\theta }_{n,\delta }(x) \in (\theta _0(x)-\delta ,\theta _0(x)+\delta )$ where $\widetilde{\Delta }_\alpha (\theta ; {\widehat{c}}_n)$ achieves a local minimum, so $\widetilde{\Delta }_\alpha ^\prime (\widehat{\theta }_{n,\delta }(x); {\widehat{c}}_n)=0$. By the first part of the proof of the theorem, ${\mathbb {P}}_{\theta _0(x)}(S_n(\delta )) \rightarrow 1$ for any $\delta$ small enough, and hence there exists a sequence $\delta _n \downarrow 0$ such that ${\mathbb {P}}_{\theta _0(x)}(S_n(\delta _n)) \rightarrow 1$ as $n \rightarrow \infty$. Now, let $\widehat{\theta }_n^*(x) = \widehat{\theta }_{n,\delta _n}(x)$ if ${\varvec{v}} \in S_n(\delta _n)$ and arbitrary otherwise. Since ${\varvec{v}} \in S_n(\delta _n)$ implies $\widetilde{\Delta }_\alpha ^\prime (\widehat{\theta }_n^*(x); {\widehat{c}}_n)=0,$ we have that

$$\begin{aligned} {\mathbb {P}}_{\theta _0(x)}(\widetilde{\Delta }_\alpha ^\prime (\widehat{\theta }_n^*(x); {\widehat{c}}_n)=0) \ge {\mathbb {P}}_{\theta _0(x)}(S_n(\delta _n)) \rightarrow 1, \end{aligned}$$

as $n \rightarrow \infty$, which establishes the existence part. For the consistency of the solution sequence, note that for any fixed $\delta > 0$ and $n$ large enough

$$\begin{aligned} {\mathbb {P}}_{\theta _0(x)}(| \widehat{\theta }_n^*(x)-\theta _0(x)| < \delta ) \ge {\mathbb {P}}_{\theta _0(x)}(| \widehat{\theta }_n^*(x)-\theta _0(x)| < \delta _n) \ge {\mathbb {P}}_{\theta _0(x)}(S_n(\delta _n)) \rightarrow 1, \end{aligned}$$

as $n\rightarrow \infty$, whence the consistency of the estimator sequence. $\square$

1.5 Proof of Theorem 2

Let $r_n:=\sqrt{nh^p{\overline{F}}(u_n;x)}$. To prove the theorem we will make use of the Cramér-Wold device (see e.g., Severini 2005, p. 337) according to which it is sufficient to show that

$$\begin{aligned} \Lambda _n := \xi ^\prime r_n [{\mathbb {T}}_n - {\mathbb {E}} ({\mathbb {T}}_n)] \leadsto N \left( 0,\frac{1}{f(x)} \;\xi ^\prime \Sigma \xi \right) , \end{aligned}$$

for all $\xi \in {\mathbb {R}}^J$.

Take an arbitrary $\xi \in {\mathbb {R}}^J$. A straightforward rearrangement of terms leads to

$$\begin{aligned} \Lambda _n&= \sum _{i=1}^n \sqrt{\frac{h^p}{n{\overline{F}}(u_n;x)}}\frac{1}{f(x)}\\&\times \, \left[ \sum _{j=1}^J \xi _j c_n^{r_j}K_{j,h}(x-X_i) \hbox {e}^{-c_n\alpha _j\left[ \left( {Y_i \over u_n}\right) ^{1/\theta _0(x)}-1\right] }\right. \left( {Y_i \over u_n}\right) ^{\beta _j} \left( \ln {Y_i \over u_n}\right) ^{r_j}_+ 1\!\!1_{\lbrace Y_i > u_n\rbrace } \\&\left. {-} {\mathbb {E}} \left( \sum _{j=1}^J \xi _j c_n^{r_j}K_{j,h}(x{-}X_i)\hbox {e}^{{-}c_n\alpha _j\left[ \left( {Y_i \over u_n}\right) ^{1/\theta _0(x)}{-}1\right] } \left( {Y_i \over u_n}\right) ^{\beta _j}\right. \right. \\&\left. \left. \times \left( \ln {Y_i \over u_n}\right) ^{r_j}_{+} 1\!\!1_{\lbrace Y_i > u_n \rbrace } \right) \right] =: \sum _{i=1}^n W_i. \end{aligned}$$

By the model assumptions, $W_1,\ldots ,W_n$ are i.i.d. random variables, and therefore ${\mathbb {V}} \text{ ar }(\Lambda _n)=n{\mathbb {V}} \text{ ar }(W_1)$. We have

$$\begin{aligned} {\mathbb {V}} \text{ ar }(W_1) = \frac{h^p}{n{\overline{F}}(u_n;x)f^2(x)} \sum _{j=1}^J \sum _{k=1}^J \xi _j\xi _k c_n^{r_j+r_k}{\mathbb {C}}_{j,k}, \end{aligned}$$

with

$$\begin{aligned} {\mathbb {C}}_{j,k}&:= {\mathbb {E}} \left[ K_{j,h}(x-X_1)K_{k,h}(x-X_1) \hbox {e}^{-c_n(\alpha _j+\alpha _k)\left[ \left( {Y_1 \over u_n}\right) ^{1/\theta _0(x)}-1\right] } \right. \\&\left. \times \left( {Y_1 \over u_n}\right) ^{\beta _j+\beta _k} \left( \ln {Y_1 \over u_n}\right) ^{r_j+r_k}_+ 1\!\!1_{\lbrace Y_1 > u_n \rbrace } \right] \\&- {\mathbb {E}} \left[ K_{j,h}(x-X_1) \hbox {e}^{-c_n\alpha _j\left[ \left( {Y_1 \over u_n}\right) ^{1/\theta _0(x)}-1\right] } \left( {Y_1 \over u_n}\right) ^{\beta _j} \left( \ln {Y_1 \over u_n}\right) _+^{r_j} 1\!\!1_{\lbrace Y_1 > u_n \rbrace } \right] \\&\times {\mathbb {E}} \left[ K_{k,h}(x-X_1) \hbox {e}^{-c_n\alpha _k\left[ \left( {Y_1 \over u_n}\right) ^{1/\theta _0(x)}-1\right] } \left( {Y_1 \over u_n}\right) ^{\beta _k} \left( \ln {Y_1 \over u_n}\right) ^{r_k}_+ 1\!\!1_{\lbrace Y_1 > u_n \rbrace }\right] . \end{aligned}$$

By using the results of Lemmas 1 and 2, we have then

$$\begin{aligned} {\mathbb {C}}_{j,k} = \frac{{\overline{F}}(u_n;x)f(x)}{h^p c_n^{r_j+r_k}}\frac{\Vert K_jK_k\Vert _1\Gamma (1+r_j+r_k) \theta _0^{r_j+r_k}(x)}{(1+\alpha _j+\alpha _k)^{1+r_j+r_k}}\;(1+o(1)), \end{aligned}$$

which gives that ${\mathbb {V}} \text{ ar }(\Lambda _n) = 1/f(x) \xi ^\prime \Sigma \xi (1+o(1))$. To establish the convergence in distribution to a normal random variable, we have to verify the Lyapunov condition for triangular arrays of random variables (Billingsley 1995, p. 362). In the present context, this simplifies to verifying that $n {\mathbb {E}}|W_1|^3 \rightarrow 0$. We have

$$\begin{aligned}&{\mathbb {E}}|W_1|^3 \le \left( \frac{h^p}{n {\overline{F}}(u_n;x)} \right) ^{3/2} \frac{1}{f^3(x)} \\&\quad \times \left\{ {\mathbb {E}} \left[ \left( \sum _{j=1}^J |\xi _j| c_n^{r_j}K_{j,h}(x-X_1) \hbox {e}^{-c_n\alpha _j\left[ \left( {Y_1 \over u_n}\right) ^{1/\theta _0(x)}-1\right] } \left( {Y_1 \over u_n}\right) ^{\beta _j} \left( \ln {Y_1 \over u_n}\right) ^{r_j}_+ 1\!\!1_{\lbrace Y_1 > u_n \rbrace } \right) ^3 \right] \right. \\&\quad + \,3\,{\mathbb {E}} \left[ \left( \sum _{j=1}^J |\xi _j| c_n^{r_j}K_{j,h}(x-X_1) \hbox {e}^{-c_n\alpha _j\left[ \left( {Y_1 \over u_n}\right) ^{1/\theta _0(x)}-1\right] } \left( {Y_1 \over u_n}\right) ^{\beta _j} \left( \ln {Y_1 \over u_n}\right) ^{r_j}_+ 1\!\!1_{\lbrace Y_1 > u_n \rbrace } \right) ^2 \right] \\&\quad \times \,{\mathbb {E}} \left[ \sum _{j=1}^J |\xi _j| c_n^{r_j}K_{j,h}(x-X_1) \hbox {e}^{-c_n\alpha _j\left[ \left( {Y_1 \over u_n}\right) ^{1/\theta _0(x)}-1\right] } \left( {Y_1 \over u_n}\right) ^{\beta _j} \left( \ln {Y_1 \over u_n}\right) ^{r_j}_+ 1\!\!1_{\lbrace Y_1 > u_n \rbrace } \right] \\&\quad \left. +\,4 \,\left[ {\mathbb {E}} \left( \sum _{j=1}^J |\xi _j| c_n^{r_j}K_{j,h}(x-X_1) \hbox {e}^{-c_n\alpha _j\left[ \left( {Y_1 \over u_n}\right) ^{1/\theta _0(x)}-1\right] } \left( {Y_1 \over u_n}\right) ^{\beta _j} \left( \ln {Y_1 \over u_n}\right) ^{r_j}_+ 1\!\!1_{\lbrace Y_1 > u_n \rbrace } \right) \right] ^3 \right\} . \end{aligned}$$

Again, by using Lemmas 1 and 2 we obtain that

$$\begin{aligned} {\mathbb {E}}|W_1|^3 = O\left( \left( n \sqrt{nh^p {\overline{F}}(u_n;x)} \right) ^{-1} \right) , \end{aligned}$$

and hence $n \text{ E } |W_1|^3 \rightarrow 0$. $\square$

1.6 Proof of Theorem 3

Apply a Taylor series expansion to the estimating equation ${\widetilde{\Delta }}_\alpha ^\prime (\widehat{\theta }_n(x);{\widehat{c}}_n)=0$ around $\theta _0(x)$. This gives

$$\begin{aligned}&0 = {\widetilde{\Delta }}_\alpha ^\prime (\theta _0(x);{\widehat{c}}_n)+\widetilde{\Delta }_\alpha ^{\prime \prime }(\theta _0(x);{\widehat{c}}_n)(\widehat{\theta }_n(x)- \theta _0(x))\\&\quad +\,\frac{1}{2}\widetilde{\Delta }_\alpha ^{\prime \prime \prime }(\widetilde{\theta }_n(x);{\widehat{c}}_n)(\widehat{\theta }_n(x)-\theta _0(x))^2 \end{aligned}$$

where $\widetilde{\theta }_n(x)$ is a random value between $\widehat{\theta }_n(x)$ and $\theta _0(x)$. A straightforward rearrangement of the terms leads then to

$$\begin{aligned}&r_n(\widehat{\theta }_n(x)-\theta _0(x))\nonumber \\&\quad = - \frac{1}{\widetilde{\Delta }_\alpha ^{\prime \prime }(\theta _0(x);{\widehat{c}}_n)+\frac{1}{2}\widetilde{\Delta }_\alpha ^{\prime \prime \prime }(\widetilde{\theta }_n(x);{\widehat{c}}_n)(\widehat{\theta }_n(x)-\theta _0(x))}\; r_n{\widetilde{\Delta }}_\alpha ^\prime (\theta _0(x);{\widehat{c}}_n) \nonumber \\&\quad = - \frac{\theta _0^{\alpha +2}(x)(1+\alpha )^2 }{1+\alpha ^2}\; r_n {\widetilde{\Delta }}_\alpha ^\prime (\theta _0(x);{\widehat{c}}_n)(1+o_{\mathbb {P}}(1)) \end{aligned}$$

(13)

by (12), the consistency of $\widehat{\theta }_n(x)$ and the boundedness of the third derivative. Another application of Taylor’s theorem gives

$$\begin{aligned} r_n \widetilde{\Delta }_\alpha ^\prime (\theta _0(x);{\widehat{c}}_n) = r_n \widetilde{\Delta }_\alpha ^\prime (\theta _0(x); c_n)-\left. \frac{\partial }{\partial {\widehat{c}}_n}\widetilde{\Delta }_\alpha ^\prime (\theta _0(x); {\widehat{c}}_n)\right| _{\widetilde{c}_n}r_n \ln \frac{\widehat{{\overline{F}}}(u_n;x)}{{\overline{F}}(u_n;x)} \end{aligned}$$

with $\widetilde{c}_n$ being a random value between ${\widehat{c}}_n$ and $c_n$. Direct computations allow us to prove that, under our assumptions, and using the second part of Lemma 2 and arguments similar to those used in the proof of Theorem 1, we have

$$\begin{aligned} \left. \frac{\partial }{\partial {\widehat{c}}_n}\widetilde{\Delta }_\alpha ^\prime (\theta _0(x); {\widehat{c}}_n)\right| _{\widetilde{c}_n}= o_{\mathbb {P}}(1). \end{aligned}$$

In addition, by Theorem 2 in de Wet et al. (2013), we deduce that

$$\begin{aligned}&r_n \widetilde{\Delta }_\alpha ^\prime (\theta _0(x);{\widehat{c}}_n) =r_n \widetilde{\Delta }_\alpha ^\prime (\theta _0(x); c_n) + o_{\mathbb {P}}(1) \nonumber \\&\quad =-\frac{\alpha }{\theta _0^{\alpha +1}(x)(1+\alpha )}\; r_n \left[ \widetilde{T}_n(K,0,0,0;x)-1 \right] \nonumber \\&\qquad + \frac{1+\alpha }{\theta _0^{\alpha +1}(x)}\; r_n \left[ \widetilde{T}_n(K,\alpha ,\alpha (1/\theta _0(x)-1),0;x)-\frac{1}{1+\alpha } \right] \nonumber \\&\qquad -\frac{1+\alpha }{\theta _0^{\alpha +2}(x)}\; r_n \left[ \widetilde{T}_n(K,\alpha ,\alpha (1/\theta _0(x)-1){+}1/\theta _0(x),1;x)- \frac{\theta _0(x)}{(1+\alpha )^2}\right] {+}o_{\mathbb {P}}(1).\nonumber \\ \end{aligned}$$

(14)

Finally, combining (13) and (14) with Theorem 2 and the delta-method, Theorem 3 follows. $\square$