A refined Weissman estimator for extreme quantiles

Abstract

Weissman extrapolation methodology for estimating extreme quantiles from heavy-tailed distributions is based on two estimators: an order statistic to estimate an intermediate quantile and an estimator of the tail-index. The common practice is to select the same intermediate sequence for both estimators. In this work, we show how an adapted choice of two different intermediate sequences leads to a reduction of the asymptotic bias associated with the resulting refined Weissman estimator. The asymptotic normality of the latter estimator is established and a data-driven method is introduced for the practical selection of the intermediate sequences. Our approach is compared to the Weissman estimator and to six bias reduced estimators of extreme quantiles on a large scale simulation study. It appears that the refined Weissman estimator outperforms its competitors in a wide variety of situations, especially in the challenging high bias cases. Finally, an illustration on an actuarial real data set is provided.

Appendices

A. Appendix: proofs

Proof of Theorem 1

It follows the same lines as the one of de Haan and Ferreira (2007, Theorem 4.3.8). Let us consider the expansion

$$\begin{aligned} \frac{\sqrt{k'_n}}{\log d_n} \left( \frac{\hat{q}_n(\alpha _n,k_n,k'_n)}{q(\alpha _n)}-1 \right) = \frac{\sqrt{k'_n}}{\log d_n} \left( \frac{X_{n-k_n,n}\; d_n^{\hat{\gamma }_n(k'_n)}}{U(1/\alpha _n)}- 1 \right) = \frac{ T_{1,n} + T_{2,n} + T_{3,n} }{T_{0,n}}, \end{aligned}$$

where we have introduced:

$$\begin{array}{l} T_{0,n} = d_n^{-\gamma } \; \frac{U(1/\alpha _n)}{U(n/k_n)},\\ T_{1,n}= \frac{\sqrt{k'_n}}{\log d_n} \left( \frac{X_{n-k_n,n}}{U(n/k_n)}-1\right) d_n^{\hat{\gamma }_n(k'_n)-\gamma },\\ T_{2,n}= \frac{\sqrt{k'_n}}{\log d_n} \left( d_n^{\hat{\gamma }_n(k'_n)-\gamma } -1\right) ,\\ T_{3,n}= \frac{\sqrt{k'_n}}{\log d_n} (1-T_{0,n}). \end{array}$$

Let us first focus on $T_{0,n}$. From de Haan and Ferreira (2007, Theorem 2.3.9), it follows from the second-order condition (5) that, for any $\varepsilon$, $\delta >0$, there exists $t_0>1$ such that for all $t\ge t_0$ and $x\ge 1$,

$$\begin{aligned} \left| \frac{1}{A_0(t)} \left( \frac{U(tx)}{U(t)}-x^{\gamma }\right) -x^{\gamma }\frac{x^{\rho }-1}{\rho } \right| \le \varepsilon x^{\gamma +\rho +\delta }, \end{aligned}$$

where $A_0$ is asymptotically equivalent to A. Letting $x=d_n$ and $t=n/k_n$ then yields

$$\begin{aligned} \left| \frac{T_{0,n}-1}{A_0(n/k_n)} - \frac{d_n^{\rho }-1}{\rho }\right| \le \varepsilon d_n^{\rho +\delta }, \end{aligned}$$

or equivalently,

$$\begin{aligned} T_{0,n} = 1 + A_0(n/k_n) \left( \frac{d_n^{\rho }-1}{\rho } + \varepsilon R_n\right) , \end{aligned}$$

where $|R_n|\le d_n^{\rho +\delta }$. Now, writing $|A_0|(t)=t^\rho \ell (t)$, where $\ell$ is a slowly-varying function, it follows,

$$\begin{aligned} A_0(n/k_n) = A_0(n/k'_n) (k'_n/k_n)^\rho \frac{\ell (n/k_n)}{\ell (n/k'_n)}, \end{aligned}$$

as $n\rightarrow \infty$. As a consequence, we obtain

$$\begin{aligned} T_{0,n} = 1 + (k'_n/k_n)^\rho A_0(n/k'_n) \left( \frac{d_n^{\rho }-1}{\rho } + \varepsilon R_n \right) \frac{\ell (n/k_n)}{\ell (n/k'_n)}, \end{aligned}$$

and letting $\varepsilon \rightarrow 0$ yields

$$\begin{aligned} T_{0,n} = 1 + (k'_n/k_n)^\rho A_0(n/k'_n) \left( \frac{d_n^{\rho }-1}{\rho } \right) \frac{\ell (n/k_n)}{\ell (n/k'_n)}. \end{aligned}$$

(21)

Second, under the assumption $\sqrt{k'_n}(\hat{\gamma }_n(k'_n)-\gamma ) {\mathop {\longrightarrow }\limits ^{d}}\mathcal{N}(\lambda \mu , \sigma ^2)$ as $\sqrt{k'_n} A(n/k_n')\rightarrow \lambda$, we get

$$\begin{aligned} d_n^{\hat{\gamma }_n(k'_n)-\gamma } = \exp \left( \frac{(\log d_n)}{\sqrt{k'_n}} \sqrt{k'_n} (\hat{\gamma }_n(k'_n)-\gamma )\right) = \exp \left( \frac{(\log d_n)}{\sqrt{k'_n}} (\lambda \mu + \sigma \xi _n) \right) , \end{aligned}$$

where $\xi _n{\mathop {\longrightarrow }\limits ^{d}}\mathcal{N}(0,1)$. Recalling that $(\log d_n)/\sqrt{k'_n} \rightarrow 0$ as $n\rightarrow \infty$, the following first order expansion holds

$$\begin{aligned} d_n^{\hat{\gamma }_n(k'_n)-\gamma } = 1 + \frac{(\log d_n)}{\sqrt{k'_n}} (\lambda \mu + \sigma \xi _n) + O_P\left( \frac{(\log d_n)^2}{k'_n}\right) . \end{aligned}$$

(22)

In particular, $d_n^{\hat{\gamma }_n(k'_n)-\gamma }{\mathop {\longrightarrow }\limits ^{\mathbb {P}}}1$ and therefore,

$$\begin{aligned} T_{1,n} = \frac{\sqrt{k'_n/k_n}}{\log d_n} \sqrt{k_n} \left( \frac{X_{n-k_n,n}}{U(n/k_n)}-1\right) (1+o_P(1)) = \frac{\sqrt{k'_n/k_n}}{\log d_n} \gamma \xi '_n (1+o_P(1)), \end{aligned}$$

(23)

where $\xi '_n{\mathop {\longrightarrow }\limits ^{d}}\mathcal{N}(0,1)$, from de Haan and Ferreira (2007, Theorem 2.2.1). Third, it immediately follows from (22) that

$$\begin{aligned} T_{2,n} = \lambda \mu + \sigma \xi _n +O_P\left( \frac{\log d_n}{\sqrt{k'_n}}\right) . \end{aligned}$$

(24)

Finally, in view of (21) and recalling that $\sqrt{k'_n}A_0(n/k'_n)\rightarrow \lambda$, one has

$$\begin{aligned} T_{3,n} = \lambda (k'_n/k_n)^\rho \left( \frac{1-d_n^{\rho }}{\rho \log d_n}\right) \frac{\ell (n/k_n)}{\ell (n/k'_n)}(1+o(1)). \end{aligned}$$

(25)

Collecting (21), (23), (24), and (25) yields

$$\begin{aligned} \begin{aligned} \frac{\sqrt{k'_n}}{\log d_n} \left( \frac{\hat{q}_n(\alpha _n,k_n,k'_n)}{q(\alpha _n)}-1 \right)&= \lambda \mu + \lambda (k'_n/k_n)^\rho \left( \frac{1-d_n^{\rho }}{\rho \log d_n}\right) \frac{\ell (n/k_n)}{\ell (n/k'_n)}(1+o(1))\\ {}&+ \sigma \xi _n + \frac{\sqrt{k'_n/k_n}}{\log d_n} \gamma \xi '_n (1+o_P(1)) + O_P\left( \frac{\log d_n}{\sqrt{k'_n}}\right) , \end{aligned} \end{aligned}$$

(26)

since $T_{0,n}=1+o(T_{3,n})$. Besides, assumptions $(k'_n/k_n)^\rho /(\log d_n)\rightarrow c\ge 0$ and $d_n\rightarrow \infty$ as $n\rightarrow \infty$ imply

$$\begin{aligned} \frac{\sqrt{k'_n}}{\log d_n} \left( \frac{\hat{q}_n(\alpha _n,k_n,k'_n)}{q(\alpha _n)}-1 \right) {\mathop {\longrightarrow }\limits ^{d}}\mathcal{N}(\lambda (\mu +c/\rho ),\sigma ^2), \end{aligned}$$

and the result is proved. $\blacksquare$

Proof of Corollary 1

Under assumption (7), Eq. (26) in the proof of Theorem 1 can be simplified as

$$\begin{aligned} \begin{aligned} \frac{\sqrt{k'_n}}{\log d_n} \left( \frac{\hat{q}_n(\alpha _n,k_n,k'_n)}{q(\alpha _n)}-1 \right)&= \lambda \mu + \lambda (k'_n/k_n)^\rho \left( \frac{1-d_n^{\rho }}{\rho \log d_n}\right) (1+o(1))\\ {}&+ \sigma \xi _n + \frac{\sqrt{k'_n/k_n}}{\log d_n} \gamma \xi '_n (1+o_P(1))+ O_P\left( \frac{\log d_n}{\sqrt{k'_n}}\right) , \end{aligned} \end{aligned}$$

since $\ell$ is asymptotically constant. Let us moreover note that

$$\begin{aligned} \frac{\log d_n}{\sqrt{k'_n}} = o\left( \frac{\sqrt{k'_n/k_n}}{\log d_n}\right) \Longleftrightarrow \frac{\sqrt{k_n}(\log d_n)^2}{k'_n}=o(1) \Longleftrightarrow \frac{\log d_n}{\sqrt{k'_n}}=o(k_n^{-1/4}) \end{aligned}$$

and therefore

$$\begin{aligned} \frac{\sqrt{k^{\prime}_n}}{\log d_n} \left( \frac{\hat{q}_n(\alpha _n,k_n,k^{\prime}_n)}{q(\alpha _n)}-1 \right) =& \lambda \mu + \lambda (k^{\prime}_n/k_n)^\rho \left( \frac{1-d_n^{\rho }}{\rho \log d_n}\right) (1+o(1)) \\&+ \sigma \xi _n + \frac{\sqrt{k^{\prime}_n/k_n}}{\log d_n} \gamma \xi^{\prime}_n(1+o_P(1)), \end{aligned}$$

which proves the result. $\blacksquare$

Proof of Lemma 1

(i) Letting $f(x)=-\rho (\log x)/(1-x^\rho )$ for all $x\ge 1$ and $\rho <0$, from (9) one has $k_n^\star = \mu ^{1/\rho } k_n (f(d_n))^{1/\rho }$ with $d_n=k_n/(n\alpha _n)\ge 1$. First, routine calculations give:

$$\begin{aligned} \frac{\partial k_n^\star }{\partial k_n} = \mu ^{1/\rho } (f(d_n))^{1/\rho } \left( 1 + \frac{d_n}{\rho } \frac{f'(d_n)}{f(d_n)}\right) = \mu ^{1/\rho } (f(d_n))^{1/\rho } \left( \frac{1}{\rho \log d_n} + \frac{1}{1-d_n^\rho } \right) \ge 0, \end{aligned}$$

for all $d_n\ge 1$. As a conclusion, ${\partial k_n^\star }/{\partial k_n}\ge 0$ which proves that $k_n^\star$ is an increasing function of $k_n$. Second, it is easily shown that f is increasing and $f(1)=1$, leading to $f(d_n)\ge 1$ and thus $k_n^\star \le \mu ^{1/\rho }k_n$.

(ii) is a consequence of $f(x) \sim -\rho \log x$ as $x\rightarrow \infty$.

(iii) Remark that assumption $\log (n\alpha _n)/\log (k_n)\rightarrow c' \le 0$ implies that $\log d_n\sim (1-c') \log k_n$ as $n\rightarrow \infty$. The conclusion follows. $\blacksquare$

Proof of Corollary 2

It is sufficient to prove that assumptions (i) and (iii) of Theorem 1 hold true. First, Lemma 1(ii) entails that $k_n^\star /k_n\sim \tau (\log d_n)^{1/\rho }$ as $n\rightarrow \infty$. Besides, from (7), we have ${A(n/k_n^\star )}/{A(n/k_n)} \sim (k_n^\star /k_n)^{-\rho },$ so that

$$\begin{aligned} \sqrt{k_n^\star } A(n/k_n^\star ) \sim (k_n^\star /k_n)^{1/2-\rho } \sqrt{k_n} A(n/k_n) \sim \tau ^{1/2-\rho } (\log d_n)^{1/(2\rho )-1} \sqrt{k_n} A(n/k_n) \rightarrow \lambda ' \tau ^{1/2-\rho } \end{aligned}$$

as $n\rightarrow \infty$ in view of the first part of (10). Assumption (i) of Theorem 1 thus holds true with $\lambda = \lambda ' \tau ^{1/2-\rho }$. Second,

$$\begin{aligned} \frac{\log d_n}{\sqrt{k_n^\star }} = \frac{\log d_n}{\sqrt{k_n}} (k_n^\star /k_n)^{-1/2} \sim \tau ^{-1/2} \frac{(\log d_n)^{1-1/(2\rho )}}{\sqrt{k_n}} \rightarrow 0 \end{aligned}$$

(27)

as $n\rightarrow \infty$ in view of the second part of (10). Third,

$$\begin{aligned} \frac{(k_n^\star /k_n)^\rho }{\log d_n} \rightarrow \tau ^\rho \end{aligned}$$

(28)

as $n\rightarrow \infty$. Collecting (27) and (28) proves that assumption (iii) of Theorem 1 thus holds true with $c=\tau ^\rho$. $\blacksquare$

Proof of Lemma 2

Recall that, from the proof of Lemma 1(iii), $\log d_n\sim (1-c') \log k_n$ as $n\rightarrow \infty$. Let us then observe that

$$\begin{aligned} \frac{\log d_n}{1-d_n^{\hat{\rho }_n}} = \frac{\log d_n}{1 - \exp (\rho \log d_n + O_P(1))} = \frac{\log d_n}{1 - O_P(d_n^\rho )} = (\log d_n) (1 + O_P(d_n^\rho )) \end{aligned}$$

and consequently

$$\begin{aligned} \left( \frac{\log d_n}{1-d_n^{\hat{\rho }_n}}\right) ^{1/\hat{\rho }_n}&= \exp \left( \frac{\log \log d_n + O_P(d_n^\rho )}{\rho + O_P(1/\log d_n)}\right) \\ {}&=\exp \left( \frac{\log \log d_n}{\rho } + O_P\left( \frac{\log \log d_n}{\log d_n}\right) + O_P(d_n^\rho ) \right) \\ {}&= (\log d_n)^{1/\rho } (1 + o_P(1)). \end{aligned}$$

Besides, since $\hat{\rho }_n$ is a consistent estimator of $\rho$ and $\mu (\cdot )$ is continuous, it follows that $(-\hat{\rho }_n \mu (\hat{\rho }_n))^{1/\hat{\rho }_n} {\mathop {\longrightarrow }\limits ^{\mathbb {P}}}(-\rho \mu (\rho ))^{1/\rho }$ and therefore

$$\begin{aligned} \begin{aligned} k_n \left( -\hat{\rho }_n \mu (\hat{\rho }_n)\frac{\log d_n}{1-d_n^{\hat{\rho }_n}}\right) ^{1/\hat{\rho }_n}&=k_n \left( -\rho \mu (\rho )(\log d_n)\right) ^{1/\rho } (1 +o_P(1))\\ {}&=k_n \left( -\rho \mu (\rho )(1-c')(\log k_n)\right) ^{1/\rho } (1 +o_P(1))\\ {}&= k_n^\star (1+o_P(1)), \end{aligned} \end{aligned}$$

in view of Lemma 1(iii). Remarking that the right hand side term tends to infinity in probability, one immediately has

$$\begin{aligned} \hat{k}_n^\star =\left\lfloor k_n \left( -\hat{\rho }_n \mu (\hat{\rho }_n)\frac{\log d_n}{1-d_n^{\hat{\rho }_n}}\right) ^{1/\hat{\rho }_n}\right\rfloor =k_n^\star (1+o_P(1)), \end{aligned}$$

and the result is proved. $\blacksquare$

Proof of Corollary 3

The first step is to prove that

$$\begin{aligned} \frac{\sqrt{\hat{k}_n^\star }}{\log d_n} \left( \frac{\hat{q}_n(\alpha _n,k_n,\hat{k}_n^\star )}{q(\alpha _n)}-1 \right) {\mathop {\longrightarrow }\limits ^{d}}\mathcal{N}(0,\sigma ^2). \end{aligned}$$

(29)

To this end, recall that $\log (n\alpha _n)/\log (k_n)\rightarrow c'\le 0$ implies that $\log d_n\sim (1-c') \log k_n$ as $n\rightarrow \infty$ and therefore condition (10) of Corollary 2 is fulfilled under the assumptions of Corollary 3. Besides, recalling that $k_n/n\rightarrow 0$ as $n\rightarrow \infty$, Lemma 2 entails that $\hat{k}_n^\star {\mathop {\longrightarrow }\limits ^{\mathbb {P}}}\infty$ and $\hat{k}_n^\star /n {\mathop {\longrightarrow }\limits ^{\mathbb {P}}}0$. Therefore, for n large enough, $\hat{k}_n^\star < n$ almost surely. Besides, for all $m_n\in \{1,\dots ,n\}$,

$$\begin{aligned} \frac{\sqrt{\hat{k}_n^\star }}{\log d_n} \left( \frac{\hat{q}_n(\alpha _n,k_n,\hat{k}_n^\star )}{q(\alpha _n)}-1 \right) | \{\hat{k}_n^\star =m_n \} {\mathop {=}\limits ^{d}}\frac{\sqrt{m_n}}{\log d_n} \left( \frac{\hat{q}_n(\alpha _n,k_n,m_n)}{q(\alpha _n)}-1 \right) . \end{aligned}$$

By Stupfler (2019, Lemma 8), since $\hat{k}_n^\star \in \{1,\dots ,n-1\}$ and $\hat{k}_n^\star {\mathop {\longrightarrow }\limits ^{\mathbb {P}}}\infty$, it is enough to show that the desired convergence (29) holds with $\hat{q}_n(\alpha _n,k_n,\hat{k}_n^\star )$ replaced by its de-conditioned version $\hat{q}_n(\alpha _n,k_n,m_n)$. This is a direct consequence of Corollary 2. The second and final step consists in replacing $\hat{k}_n^\star$ by its non random version $k_n^\star$ in the rate of convergence of (29). This can be achieved using Lemma 2 and Slutsky’s lemma. $\blacksquare$

B. Appendix: tables

Table 3 RMSEs associated with eight estimators of the extreme quantile $q(\alpha _n=1/n)$ on a Burr distribution. The best result is emphasized in bold. RMSEs larger than 1 are not reported

Table 4 RMSEs associated with eight estimators of the extreme quantile $q(\alpha _n=1/n)$ on a NHW distribution. The best result is emphasized in bold. RMSEs larger than 1 are not reported

Table 5 RMSEs associated with eight estimators of the extreme quantile $q(\alpha _n=1/n)$ on five heavy-tailed distributions. The best result is emphasized in bold. RMSEs larger than 1 are not reported

Table 6 RMSEs associated with eight estimators of the extreme quantile $q(\alpha _n=1/(2n))$ on a Burr distribution. The best result is emphasized in bold. RMSEs larger than 1 are not reported

Table 7 RMSEs associated with eight estimators of the extreme quantile $q(\alpha _n=1/(2n))$ on a NHW distribution. The best result is emphasized in bold. RMSEs larger than 1 are not reported

Table 8 RMSEs associated with eight estimators of the extreme quantile $q(\alpha _n=1/(2n))$ on five heavy-tailed distributions. The best result is emphasized in bold. RMSEs larger than 1 are not reported