Improving precipitation forecasts using extreme quantile regression

Abstract

Aiming to estimate extreme precipitation forecast quantiles, we propose a nonparametric regression model that features a constant extreme value index. Using local linear quantile regression and an extrapolation technique from extreme value theory, we develop an estimator for conditional quantiles corresponding to extreme high probability levels. We establish uniform consistency and asymptotic normality of the estimators. In a simulation study, we examine the performance of our estimator on finite samples in comparison with a method assuming linear quantiles. On a precipitation data set in the Netherlands, these estimators have greater predictive skill compared to the upper member of ensemble forecasts provided by a numerical weather prediction model.

Appendix: Proofs

This section contains the proofs of Theorems 1–3 in Section 3. Throughout this section, c,c₁,c₂,… denote positive constants, which are not necessarily the same at each occurrence.

1.1 Proof of Theorem 1

The uniform consistency of $\hat r$ relies heavily on the uniform Bahadur representation for $\hat r$. We make use of the Bahadur representation obtained in Kong et al. (2010).

Let ψ_τ(u) = τ − I(u < 0), that is the right derivative of ρ_τ at u. Then by Corollary 3.3 and Proposition 1 in Kong et al. (2010), we have

$$ \begin{array}{@{}rcl@{}} &&\sup_{x \in [a,b]} \left| \hat{r}(x) - r(x) + {h_{n}^{2}} c r^{\prime\prime}(x) - \frac{1}{nh_{n}}\sum\limits_{i=1}^{n} \psi_{\tau_{c}}(\epsilon_{i})C_{n, i}(x)K\left( \frac{X_{i}-x}{h_{n}}\right) \right|\\ &=& O_{p}\left( \left\{\frac{\log n}{nh_{n}}\right\}^{3/4}\right) = O_{p}\left( \left\{\frac{\log n}{n^{1-\delta_{h}}}\right\}^{3/4}\right), \end{array} $$

where C_n,i(x) is a Lipschitz continuous function and thus absolutely bounded in [a,b]. Define

$$ {\Delta}_{n}(x) = \frac{1}{nh_{n}}\sum\limits_{i=1}^{n} \psi_{\tau_{c}}(\epsilon_{i})C_{n,i}(x)K\left( \frac{X_{i}-x}{h_{n}}\right). $$

Then, the triangle inequality leads to

$$ \begin{array}{@{}rcl@{}} \sup_{x \in [a,b]} \left|\hat{r}(x) - r(x) \right|&\leq&\sup_{x \in [a,b]} \left|{h_{n}^{2}} c r^{\prime\prime}(x)\right|+ \sup_{x \in [a,b]} \left|{\Delta}_{n}(x)\right|+ O_{p}\left( \left\{\frac{\log n}{n^{1-\delta_{h}}}\right\}^{3/4}\right) \\ &&=O(n^{-2\delta_{h}})+\sup_{x \in [a,b]} \left|{\Delta}_{n}(x)\right|+ O_{p}\left( \left\{\frac{\log n}{n^{1-\delta_{h}}}\right\}^{3/4}\right). \end{array} $$

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The last equality follows from the fact that r^″ is uniformly bounded by Assumption A1.

Next, we show that, there exists a $\delta _{C}\in (0, \frac {1}{2}-\delta _{h})$ such that

$$ \sup_{x \in [a,b]} \left|{\Delta}_{n}(x)\right| = o_{p}(n^{-\delta_{C}}). $$

(13)

Define $T_{i}(x):= h_{n} K\left (\frac {X_{i}-x}{h_{n}}\right )C_{n, i}(x)$. Then for any x,y ∈ [a,b], by the triangle inequality and the Lipschitz continuity of K, we have

$$ \begin{array}{@{}rcl@{}} &&\left|T_{i}(x)-T_{i}(y)\right|= h_{n} \left|K\left( \frac{X_{i}-x}{h_{n}}\right)C_{n, i}(x) - K\left( \frac{X_{i}-y}{h_{n}}\right)C_{n,i}(y)\right| \\ &&\leq h_{n} |C_{n, i}(x)|\left|K\left( \frac{X_{i}-x}{h_{n}}\right) - K\left( \frac{X_{i}-y}{h_{n}}\right)\right| + h_{n}K\left( \frac{X_{i}-y}{h_{n}}\right)|C_{n, i}(x) - C_{n,i}(y)|\\ &&\leq c_{1}\left|x-y\right| + c_{2} h_{n} |x-y|\sup_{u\in [-1,1]}K(u)\\ &&\leq c|x-y|. \end{array} $$

Note that the constant c does not depend on i, that is, the Lipschitz continuity is uniform in i for all T_i’s. Consequently, it follows from that |ψ_τ(u)| ≤ 1 that,

$$ \begin{array}{@{}rcl@{}} |{\Delta}_{n}(x) - {\Delta}_{n}(y)| = \frac{1}{n{h^{2}_{n}}}\left|\sum\limits_{i=1}^{n} \psi_{\tau_{c}}(\epsilon_{i})(T_{i}(x) - T_{i}(y)) \right| \leq c\frac{|x-y|}{{h_{n}^{2}}}. \end{array} $$

Let $M_{n} = n^{\delta _{C}+2\delta _{h}}\log n$ and {I_i = (t_i,t_i+ 1],i = 1,…,M_n} be a partition of (a,b], where $t_{i+1} - t_{i} = \frac {b-a}{M_{n}}$. Then for t ∈ I_i,

$$ \begin{array}{@{}rcl@{}} |{\Delta}_{n}(t) - {\Delta}_{n}(t_{i})| \leq \frac{c(b-a)}{M_{n} {h_{n}^{2}}}, \end{array} $$

or equivalently,

$$ \begin{array}{@{}rcl@{}} {\Delta}_{n}(t_{i}) - \frac{c(b-a)}{M_{n} {h_{n}^{2}}} \leq {\Delta}_{n}(t) \leq {\Delta}_{n}(t_{i}) + \frac{c(b-a)}{M_{n} {h_{n}^{2}}}. \end{array} $$

Therefore, for n sufficiently large,

$$ \begin{array}{@{}rcl@{}} \mathrm{P}\left( \sup_{x\in[a,b]} |{\Delta}_{n}(x)| > n^{-\delta_{C}} \right) &=& \mathrm{P}\left( \max_{1\leq i \leq M_{n}} \sup_{t\in I_{i}}|{\Delta}_{n}(t)| > n^{-\delta_{C}} \right)\\ &&\leq \sum\limits_{i=1}^{M_{n}} \mathrm{P}\left( \sup_{t\in I_{i}} |{\Delta}_{n}(t)| > n^{-\delta_{C}} \right)\\&& \leq \sum\limits_{i=1}^{M_{n}} \mathrm{P}\left( |{\Delta}_{n}(t_{i})| > n^{-\delta_{C}} - \frac{c(b-a)}{M_{n} {h_{n}^{2}}}\right)\\ &&\leq\sum\limits_{i=1}^{M_{n}} \mathrm{P}\left( |{\Delta}_{n}(t_{i})| > \frac{1}{2}n^{-\delta_{C}} \right)\\ &=&\sum\limits_{i=1}^{M_{n}} \mathrm{P}\left( \left| \sum\limits_{j=1}^{n} \frac{T_{j}(t_{i})\psi_{\tau_{c}}(\epsilon_{j})}{h_{n}} \right| > \frac{1}{2}h_{n}n^{1-\delta_{C}} \right) =: \sum\limits_{i=1}^{M_{n}} P_{i}, \end{array} $$

where the third inequality is due to that $\frac {c(b-a)}{M_{n} {h_{n}^{2}}}<\frac {1}{2}n^{-\delta _{C}}$ for n sufficiently large. Next, we apply Hoeffding’s inequality to bound P_i. Define

$$ W_{n,i,j}:= \frac{T_{j}(t_{i})\psi_{\tau_{c}}(\epsilon_{j})}{h_{n}}= K\left( \frac{X_{j}-t_{i}}{h_{n}}\right)C_{n, j}(t_{i}) \psi_{\tau_{c}}(\epsilon_{j}). $$

For each i and n, {W_n,i,j, 1 ≤ j ≤ n} is a sequence of i.i.d. random variables. And with probability one, |W_n,j,i|≤ sup− 1≤u≤ 1K(u) supa≤x≤bC_n,i(x) =: c₃. Moreover, $\mathbb {E}\left (W_{n,j,i} \right )=0$ because $\mathbb {E}(\psi _{\tau _{c}}(\epsilon _{j}))=0$ and X_j and 𝜖_j are independent. Thus, by Hoeffding’s inequality,

$$ \begin{array}{@{}rcl@{}} P_{i}= \mathrm{P}\left( \left| {\sum}_{j=1}^{n} W_{n,i,j}\right| \geq \frac{1}{2}h_{n}n^{1-\delta_{C}}\right)\leq 2\exp\left( -\frac{n^{1-2\delta_{C}} {h_{n}^{2}}}{8{c_{3}^{2}}} \right)=2\exp\left( -cn^{1- 2 \delta_{h} -2\delta_{C}}\right). \end{array} $$

Note that 1 − 2δ_h − 2δ_C > 0 by the choice of δ_C. Thus, for n →∞,

$$ \mathrm{P}\left( \sup_{x\in[a,b]} |{\Delta}_{n}(x)| > n^{-\delta_{C}}\right) \leq 2M_{n}\exp\left( -cn^{1-2\delta_{h}-2\delta_{C}}\right)\rightarrow 0. $$

Hence, Eq. 2 is proved. Now by choosing δ = δ_C, we obtain via (1) that,

$$ n^{\delta} \sup_{x\in [a,b]} | \hat{r}_{n}(x) - r(x)| = O(n^{\delta_{C}-2\delta_{h}}) + o_{p}(1) + O_{p}\left( n^{-\frac{3}{4} + \frac{3}{4}\delta_{h} +\delta_{C}}(\log n)^{\frac{3}{4}}\right) = o_{p}(1), $$

due to that $\delta _{h}\in (\frac {1}{5}, \frac {1}{2})$ and $\delta _{C}<\frac {1}{2}-\delta _{h}$.

1.2 Proof of Theorem 2

The proof follows a similar line of reasoning as that of Theorem 2.1 in Wang et al. (2012). The uniform consistency of $\hat r_{n}$ given in Theorem 1 plays a crucial role. Define $V_{n} := ||\hat {r}_{n}-r||_{\infty } = o_{p}\left (n^{-\delta }\right )$.

Let U_i = F_{Y |X}(Y_i|X_i) for all 1 ≤ i ≤ n. Then {U_i,i = 1,…,n} constitute i.i.d. random variables from a standard uniform distribution. Recall the definition of e_i:

$$ e_{i}=Y_{i}-\hat r_{n}(X_{i})=Q_{Y|X}(U_{i}|X_{i})-\hat r_{n}(X_{i}). $$

Thus, the ordering of {e_i,i = 1,…,n} is not necessarily the same as the ordering of {U_i,i = 1,…,n}. The main task of this proof is to show that the k_n largest e_i’s correspond to the k_n largest U_i’s; see Eq. 4. To this aim, we first prove that with probability tending to one, e_n−j,n for j = 0,…,k_n can be decomposed as follows,

$$ e_{n-j,n} = Q_{\epsilon}(U_{i(j)}) + r(X_{i(j)})-\hat r_{n}(X_{i(j)}) \text{ for } j=0,{\ldots} k_{n} , $$

(14)

where i(j) is the index function defined as e_i(j) = e_n−j,n. In view of Eq. ??, it is sufficient to prove that with probability tending to one, U_i(j) > τ_c jointly for all j = 0,…,k_n. Define another index function, $\tilde {i}(j)$ by $U_{\tilde {i}(j)} = U_{n-j,n}$. Then it follows for n large enough,

$$ \begin{array}{@{}rcl@{}} \mathrm{P} \left( \cup_{j=0}^{k_{n}} \{ U_{i(j)} < \tau_{c} \}\right) &=& \mathrm{P}\left( \cup_{j=0}^{k_{n}} \{ Y_{i(j)} < Q_{Y|X}(\tau_{c}|X_{i(j)})\} \right)\\ &=& \mathrm{P}\left( \min_{0\leq j \leq k_{n}} \left( Y_{i(j)} - r(X_{i(j)})\right) < 0 \right)\\ &=& \mathrm{P}\left( \min_{0\leq j \leq k_{n}}\left( Y_{i(j)} - \hat{r}_{n}(X_{i(j)}) - r(X_{i(j)}) + \hat{r}_{n}(X_{i(j)})\right) < 0 \right)\\ &&\leq \mathrm{P}\left( \min_{0\leq j \leq k_{n}} e_{n-j,n} - \sup_{x \in [a,b]}|\hat{r}_{n}(x) - r(x)| < 0 \right)\\ &=& \mathrm{P}\left( e_{n-k_{n},n} < V_{n} \right) = 1-\mathrm{P}(e_{n-k_{n},n}\geq V_{n})\\ &&\leq 1- \mathrm{P}\left( \cap_{j=0}^{k_{n}} \{ e_{\tilde{i}(j)} \geq V_{n} \} \right)\\ &=& 1- \mathrm{P}\left( \cap_{j=0}^{k_{n}} \left\{ Q_{\epsilon}(U_{n-j,n}) + r(X_{\tilde{i}(j)}) - \hat{r}_{n}(X_{\tilde{i}(j)}) \geq V_{n} \right\} \right)\\ &&\leq 1- \mathrm{P}\left( Q_{\epsilon}(U_{n-k_{n},n}) \geq 2V_{n} \right), \end{array} $$

where the second equality follows from that Q_{Y |X}(τ_c|X_i(j)) = r(X_i(j)) and the last equality follows from (??) and the fact that $U_{n-k_{n},n} > \tau _{c}$ for n large enough. Then, $\lim _{n\to \infty } \mathrm {P} \left (\cup _{j=0}^{k_{n}} \{U_{i(j)} < \tau _{c} \}\right ) = 0$ follows from $Q_{\epsilon }(U_{n-k_{n},n}) \to \infty$ and V_n = o_p(1) as n →∞. Hence, Eq. 3 is proved.

Next, we show that

$$ \lim_{n\rightarrow\infty}\mathrm{P}\left( \cap_{j=0}^{k_{n}} \{e_{n-j,n}= Q_{\epsilon}(U_{n-j,n})+ r(X_{i(j)})-\hat r_{n}(X_{i(j)})\}\right)=1, $$

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that is the ordering of k largest residuals is determined by the ordering of U_i’s. In view of Eq. 3, it is sufficient to show that with probability tending to one,

$$ \min_{1 \leq i \leq k_{n}} (Q_{\epsilon}(U_{n-i+1,n})-Q_{\epsilon}(U_{n-i,n}))\geq 2\max_{1 \leq i \leq k_{n}}|r(X_{i(j)})-\hat r_{n}(X_{i(j)}|. $$

(16)

By the second order condition given in Eq. ?? and Theorem 2.3.9 in De Haan and Ferreira (2007), for any small δ₁,δ₂ > 0, and n large enough,

$$ \frac{Q_{\epsilon}(U_{n-i+1,n})} {Q_{\epsilon}(U_{n-i,n})} \geq W_{i}^{\gamma}+A_{0}\left( \frac{1}{1-U_{n-i,n}}\right)W_{i}^{\gamma}\frac{W_{i}^{\rho}-1}{\rho}-\delta_{1}\left|A_{0}\left( \frac{1}{1-U_{n-i,n}}\right)\right|W_{i}^{\gamma+\rho+\delta_{2}}, $$

(17)

for i = 1,…,k_n, where $W_{i}=\frac {1-U_{n-i,n}}{1-U_{n-i+1,n}}$ and $\lim _{t\rightarrow \infty }A_{0}(t)/A(t)=1$. Observe that $\log W_{i}= \log \frac {1}{1-U_{n-i+1,n}}-\log \frac {1}{1-U_{n-i,n}} \overset {d}{=}E_{n-i+1,n}-E_{n-i,n}$ with E_i’s i.i.d. standard exponential variables. Thus, by Rènyi’s representation (Rényi 1953), we have

$$ \{W_{i}, 1\leq i\leq k_{n}\}\overset{d}{=}\left\{\exp\left( \frac{E_{i}}{i}\right), 1\leq i\leq k_{n} \right\}. $$

From Proposition 2.4.9 in De Haan and Ferreira (2007), we have $\frac {U_{n-k_{n},n}}{1-\frac {k_{n}}{n}}\overset {P}{\rightarrow }1$, which implies that ${A_{0}\left (\frac {1}{1-U_{n-k_{n},n}}\right )}=O_{p}\left ({A_{0}\left (\frac {n}{k_{n}}\right )}\right )$. Using the fact that A₀ is regularly varying with index $\mathcal {\rho }$, hence |A₀| is ultimately decreasing, we obtain for n sufficiently large and any i = 1,…,k_n,

$$ \begin{array}{@{}rcl@{}} \left|A_{0}\left( \frac{1}{1-U_{n-i,n}}\right)\right|&\leq& \left|A_{0}\left( \frac{1}{1-U_{n-k_{n},n}}\right)\right| \\ &=&\left|O_{p}\left( A_{0}\left( \frac{n}{k_{n}}\right)\right)\right|=\left|O_{p}\left( A\left( \frac{n}{k_{n}}\right)\right)\right|=\left|O_{p}\left( \frac{1}{\sqrt{k_{n}}}\right)\right|, \end{array} $$

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by the assumption $\sqrt {k_{n}}A\left (\frac {n}{k_{n}}\right )\rightarrow \lambda$.

For a sufficiently large u and any k_n ≥ 1,

$$ \begin{array}{@{}rcl@{}} &&\mathrm{P}\left( \max_{1\leq i\leq k_{n}}\frac{E_{i}}{i}\leq u\right)=\prod\limits_{i=1}^{k_{n}} \left( 1-e^{-iu}\right)=\exp\left( \sum\limits_{i=1}^{k_{n}}\log \left( 1-e^{-iu}\right) \right)\\ &=&\exp\left( -\sum\limits_{i=1}^{k_{n}}\sum\limits_{j=1}^{\infty} j^{-1}e^{-iuj} \right)\geq \exp\left( -\sum\limits_{i=1}^{k_{n}}e^{-iu}\right)= \exp\left( \frac{1-e^{-ku}}{1-e^{u}}\right), \end{array} $$

which tends to one as u →∞. This implies that

$$ \min_{1\leq i\leq k_{n}}W_{i}^{\rho}\overset{d}{=}\exp\left( \rho \max_{1\leq i\leq k_{n}}\frac{E_{i}}{i}\right)=O_{p}(1). $$

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Thus, combining Eqs. 6, 7 and 8, we have

$$ \begin{array}{@{}rcl@{}} &&\min_{1\leq i\leq k_{n}}\frac{Q_{\epsilon}(U_{n-i+1,n})} {Q_{\epsilon}(U_{n-i,n})}-1 \\ &\geq& \min_{1\leq i\leq k_{n}}W_{i}^{\gamma}\left( 1-\left|O_{p}\left( \frac{1}{\sqrt{k_{n}}}\right) \right| \left( \frac{W_{i}^{\rho}-1}{\rho}+ \delta_{1}W_{i}^{\rho+\delta_{2}} \right)\right)-1\\ &=&\min_{1\leq i\leq k_{n}}W_{i}^{\gamma}\left( 1-\left|O_{p}\left( \frac{1}{\sqrt{k_{n}}}\right)\right|\right)-1 \overset{d}{=}\exp\left( \gamma\frac{E_{1}}{k_{n}}\right)\left( 1-\left|O_{p}\left( \frac{1}{\sqrt{k_{n}}}\right)\right|\right)-1\\ &=&\frac{\gamma E_{1}}{k_{n}}\left( 1-\left|O_{p}\left( \frac{1}{\sqrt{k_{n}}}\right)\right|\right), \end{array} $$

where the third equality follows from that $\min _{1\leq i\leq k_{n}}\frac {E_{i}}{i}\overset {d}{=}E_{1,k}\overset {d}{=}\frac {E_{1}}{k}$ by Rènyi’s representation. Thus, we obtain that

$$ \begin{array}{@{}rcl@{}} \min_{1 \leq i \leq k_{n}} (Q_{\epsilon}(U_{n-i+1,n})-Q_{\epsilon}(U_{n-i,n})) &\geq& \left( Q_{\epsilon}(U_{n-k_{n},n})\frac{\gamma E_{1}}{k_{n}}\right)\left( 1\!-\left|O_{p}\left( \frac{1}{\sqrt{k_{n}}}\right)\right|\right) \\ &=& \left( \frac{n}{k_{n}}\right)^{\gamma} k_{n}^{-1}|O_{p}(1)|. \end{array} $$

Thus, Eq. 5 is proved by the assumption $k_{n}^{-1}\left (\frac {n}{k_{n}}\right )^{\gamma }>>n^{-\delta }$ and $\max _{1 \leq i \leq k_{n}}|r(X_{i(j)})-\hat r_{n}(X_{i(j)}|\leq 2 V_{n}=o_{p}\left (n^{-\delta }\right )$. Intuitively, Eq. 5 means that the difference between two successive upper order statistics of 𝜖 is larger than the error made in the estimation of r(x).

As aforementioned, Eqs. 3 and 5 together lead to Eq. 4, which further implies that with probability tending to one,

$$ \max_{0 \leq j \leq k_{n}} \left|\frac{e_{n-j,n}}{Q_{\epsilon}(U_{n-j,n})}-1 \right| \leq \frac{V_{n}}{Q_{\epsilon}(U_{n-k_{n},n})} =o_{p}\left( n^{-\delta}\left( \frac{k_{n}}{n}\right)^{\gamma}\right). $$

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By the definition of $\hat {\gamma }_{n}$ and Eq. 9, we can write the estimator as follows,

$$ \begin{array}{@{}rcl@{}} \hat{\gamma}_{n} &=& \frac{1}{k_{n}} \sum\limits_{i=0}^{k_{n}-1}\log \frac{e_{n-i,n}}{e_{n-k_{n},n}}\\ &=& \frac{1}{k_{n}} \sum\limits_{i=0}^{k_{n}-1} \log\frac{Q_{\epsilon}(U_{n-i,n})}{Q_{\epsilon}(U_{n-k_{n},n})} + \left( \frac{1}{k_{n}} \sum\limits_{i=0}^{k_{n}-1}\log \frac{e_{n-i,n}}{Q_{\epsilon}(U_{n-i,n})}-\log \frac{e_{n-k_{n},n}}{Q_{\epsilon}(U_{n-k_{n},n})}\right)\\ &=:& \hat{\gamma}_{H} + o_{p}\left( n^{-\delta}\left( \frac{k_{n}}{n}\right)^{\gamma}\right). \end{array} $$

The first part is the well known Hill estimator and we have by Theorem 3.2.5 in De Haan and Ferreira (2007),

$$ \sqrt{k_{n}}(\hat{\gamma}_{H}-\gamma)) \xrightarrow{d} N\left( \frac{\lambda}{1-\rho},\gamma^{2}\right). $$

Therefore we can conclude,

$$ \sqrt{k_{n}}(\hat{\gamma}_{n} - \gamma) = \sqrt{k_{n}}(\hat{\gamma}_{H}-\gamma) +o_{p}\left( \sqrt{k_{n}}n^{-\delta}\left( \frac{k_{n}}{n}\right)^{\gamma}\right) \xrightarrow{d} N\left( \frac{\lambda}{1-\rho},\gamma^{2}\right), $$

by the assumption that $k_{n}^{\gamma +1}n^{-\gamma -\delta }\rightarrow 0$.

We remark that the proof for Theorem 2.1 in Wang et al. (2012) isn’t completely rigorous, namely, the proof for (S.1) in the supplementary material of that paper is not right. We fix the problem while proving (9), which is an analogue to (S.1).

1.3 Proof of Theorem 4

Before we proceed with the proof of Theorem 3, we state the asymptotic normality of $\hat {Q}_{\epsilon }(\tau _{n})$ defined in Eq. ?? in the theorem below.

Theorem 1

Let the conditions of Theorem 2 be satisfied. Assumenp_n = o(k_n) and$\log (np_{n}) = o(\sqrt {k_{n}})$,then, asn →∞,

$$ \frac{\sqrt{k_{n}}}{\log(k_{n}/(np_{n}))} \left( \frac{\hat{Q}_{\epsilon}(\tau_{n})}{Q_{\epsilon}(\tau_{n})} -1 \right) \xrightarrow{d} N\left( \frac{\lambda}{1-\rho},\gamma^{2}\right). $$

(21)

Theorem 4 can be proved in the same way as that for Theorem 2 in Wang et al. (2012). For the sake of completeness, we present the proof in this section.

Recall that $\hat {Q}_{\epsilon }(\tau _{n}) = \left (\frac {k_{n}}{np_{n}} \right )^{\hat {\gamma }_{n}} e_{n-k_{n},n}=:d_{n}^{\hat {\gamma }_{n}} e_{n-k_{n},n}$. First, note that from Theorem 2, we have $\sqrt {k_{n}}(\hat {\gamma }_{n}-\gamma )=\Gamma +o_{p}(1)$, where Γ is a random variable from $N\left (\frac {\lambda }{1-\rho },\gamma ^{2}\right )$. Therefore,

$$ \begin{array}{@{}rcl@{}} d_{n}^{\hat{\gamma}_{n}-\gamma}&=&\exp\left( (\hat{\gamma}_{n}-\gamma)\log d_{n}\right)=\exp\left( \frac{\log d_{n}}{\sqrt{k_{n}}} (\Gamma+o_{p}(1)) \right) \\ &=&1+ \frac{\log d_{n}}{\sqrt{k_{n}}}\Gamma+o_{p}(\frac{\log d_{n}}{\sqrt{k_{n}}}), \end{array} $$

(22)

where the last step follows from the assumption that $\frac {\log d_{n}}{\sqrt {k_{n}}} \to 0$. Second, by Theorem 2.4.1,

$$ \sqrt{k}\left( \frac{Q_{\epsilon}(U_{n-k_{n},n})} {Q_{\epsilon}(1-k_{n}/n)}-1\right)\xrightarrow{d} N(0,\gamma^{2}). $$

In combination with Eq. 9, we have

$$ \begin{array}{@{}rcl@{}} \frac{e_{n-k_{n},n}}{Q_{\epsilon}(1-k_{n}/n)}&=&\frac{e_{n-k_{n},n}}{Q_{\epsilon}(U_{n-k_{n},n})} \cdot\frac{Q_{\epsilon}(U_{n-k_{n},n})}{Q_{\epsilon}(1-k_{n}/n)} =\left( 1+o_{p}\left( n^{-\delta}\left( \frac{k_{n}}{n}\right)^{\gamma}\right)\right)\left( 1+O_{p}\left( \frac{1}{\sqrt{k_{n}}}\right)\right) \\ &=&1+O_{p}\left( \frac{1}{\sqrt{k_{n}}}\right), \end{array} $$

(23)

by the assumption that $k_{n}^{\gamma +1}n^{-\gamma -\delta }\rightarrow 0$. Last, by the second order condition given in Eq. 7 and Theorem 2.3.9 in De Haan and Ferreira (2007),

$$ \frac{Q_{\epsilon}(1-p_{n})}{Q_{\epsilon}(1-k_{n}/n) d_{n}^{\gamma}}=1+O(A(n/k_{n}))=1+O\left( \frac{1}{\sqrt{k_{n}}}\right). $$

(24)

Finally, combing (22), (23) and (24), we have

$$ \begin{array}{@{}rcl@{}} \frac{\hat{Q}_{\epsilon}(\tau_{n})}{Q_{\epsilon}(\tau_{n})} &=&\frac{d_{n}^{\hat{\gamma}}e_{n-k_{n},n}}{Q_{\epsilon}(1-p_{n})} =d_{n}^{\hat{\gamma}_{n}-\gamma}\frac{e_{n-k_{n},n}}{Q_{\epsilon}(1-k_{n}/n)}\cdot \frac{Q_{\epsilon}(1-k_{n}/n) d_{n}^{\gamma}}{Q_{\epsilon}(1-p_{n})}\\ &=&\left( 1+ \frac{\log d_{n}}{\sqrt{k_{n}}}\Gamma+o_{p}\left( \frac{\log d_{n}}{\sqrt{k_{n}}}\right)\right)\left( 1+O_{p}\left( \frac{1}{\sqrt{k_{n}}}\right) \right)\left( 1+O\left( \frac{1}{\sqrt{k_{n}}}\right)\right)\\ &=&1+ \frac{\log d_{n}}{\sqrt{k_{n}}}\Gamma+o_{p}\left( \frac{\log d_{n}}{\sqrt{k_{n}}} \right), \end{array} $$

by the assumption that d_n →∞. Thus, Eq. 21 follows immediately.

1.4 Proof of Theorem 3

By definition of $\hat {Q}_{Y|X}(\tau _{n}|x)$ and Theorem 1, we have,

$$ \begin{array}{@{}rcl@{}} &&\frac{\sqrt{k_{n}}}{\log\left( \frac{k_{n}}{np_{n}}\right)Q_{\epsilon}(\tau_{n})}\left( \hat{Q}_{Y|X}(\tau_{n}|x) - Q_{Y|X}(\tau_{n}|x)\right)\\ &=&\frac{\sqrt{k_{n}}}{\log\left( \frac{k_{n}}{np_{n}}\right)Q_{\epsilon}(\tau_{n})} \left( \hat{Q}_{\epsilon}(\tau_{n}) - {Q}_{\epsilon}(\tau_{n})+\hat r_{n}(x)-r(x) \right),\\ &=&\frac{\sqrt{k_{n}}}{\log\left( \frac{k_{n}}{np_{n}}\right)Q_{\epsilon}(\tau_{n})} \left( \hat{Q}_{\epsilon}(\tau_{n}) - {Q}_{\epsilon}(\tau_{n})\right)+O_{p}\left( \frac{\sqrt{k_{n}}n^{-\delta}}{\log\left( \frac{k_{n}}{np_{n}}p_{n}^{-\gamma}\right)}\right). \end{array} $$

Thus it follows from Theorem 4 and the assumption $\frac {\sqrt {k}p_{n}^{\gamma }}{n^{\delta }\log \left (\frac {k_{n}}{np_{n}}\right )} \to 0$ that

$$ \frac{\sqrt{k_{n}}}{\log\left( \frac{k_{n}}{np_{n}}\right)Q_{\epsilon}(\tau_{n})}\left( \hat{Q}_{Y|X}(\tau_{n}|x) - Q_{Y|X}(\tau_{n}|x)\right) \xrightarrow{d} N\left( \frac{\lambda}{1-\rho},\gamma^{2}\right). $$