Methods for estimating the upcrossings index: improvements and comparison

Abstract

The upcrossings index $0\le \eta \le 1,$ as a measure of the degree of local dependence in the upcrossings of a high level by a stationary process, plays, together with the extremal index $\theta ,$ an important role in extreme events modelling. For stationary processes, verifying a long range dependence condition, upcrossings of high thresholds in different blocks can be assumed asymptotically independent and therefore blocks estimators for the upcrossings index can be easily constructed using disjoint blocks. In this paper we focus on the estimation of the upcrossings index via the blocks method and properties such as consistency and asymptotic normality are studied. Besides this new estimation approach for this parameter, we also enlarge its family of runs estimators and improve estimation within this class by providing an empirical way of checking local dependence conditions that control the clustering of upcrossings. We compare the performance of a range of different estimators for $\eta$ and illustrate the methods using simulated data and financial data.

Appendix A: Proofs for section 2

1.1 Proof of Theorem 1

Let us consider the sample $X_1,\ldots , X_{[n/c_n]},$ divided into $k_n/c_n$ disjoint blocks. We can then apply the arguments used in Lemma 2.1 of Ferreira (2006) and conclude that

$$\begin{aligned} P(\widetilde{N}_{[n/c_n]}(v_n)=0)\sim P^{k_n/c_n}(\widetilde{N}_{r_n}(v_n)=0). \end{aligned}$$

(5.1)

Now, from the definition of the upcrossings index $\eta$ we have $P(\widetilde{N}_{[n/c_n]}(v_n)=0)\xrightarrow [n\rightarrow +\infty ]{} e^{-\eta \nu },$ hence applying logarithms on both sides of (5.1), (2.4) follows immediately.

For the conditional mean number of upcrossings in each block, we have from (2.4) and the definition of the thresholds $v_n$

$$\begin{aligned} E[\widetilde{N}_{r_n}(v_n)\ |\ \widetilde{N}_{r_n}(v_n)>0]= & {} \frac{\sum _{j\ge 1} j\times P(\widetilde{N}_{r_n}(v_n)=j,\ \widetilde{N}_{r_n}(v_n)>0)}{P(\widetilde{N}_{r_n}(v_n)>0)}\\= & {} \frac{E[\widetilde{N}_{r_n}(v_n)]}{P(\widetilde{N}_{r_n}(v_n)>0)}=\frac{r_nP(X_1\le v_n<X_2)}{P(\widetilde{N}_{r_n}(v_n)>0)}\\\sim & {} \frac{r_nc_n\nu /n}{c_n\eta \nu /k_n} , \end{aligned}$$

which completes the proof since $k_nr_n\sim n.$$\square$

1.2 Proof of Theorem 2

It suffices to show that

$$\begin{aligned} \lim _{n\rightarrow +\infty } \sum _{j=1}^{+\infty } j\widetilde{\pi }_n(j;v_n)=\eta ^{-1} \end{aligned}$$

(5.2)

and

$$\begin{aligned} \sum _{j=1}^{+\infty }j\big (\widehat{\widetilde{\pi }}_n(j;v_n)-\widetilde{\pi }_n(j;v_n)\big )\xrightarrow [n\rightarrow +\infty ]{P}0. \end{aligned}$$

(5.3)

(5.2) follows immediately from Theorem 1 since

$$\begin{aligned} \lim _{n\rightarrow +\infty } \sum _{j=1}^{+\infty } j\widetilde{\pi }_n(j;v_n)=\lim _{n\rightarrow +\infty }E\big [\widetilde{N}_{r_n}(v_n)\ |\ \widetilde{N}_{r_n}(v_n)>0\big ]=\eta ^{-1}. \end{aligned}$$

To show (5.3), lets start by noting that Theorem 2 of Hsing (1991) holds for $\widehat{\widetilde{\pi }}_n(j;v_n)$ and $\widetilde{\pi }_n(j;v_n),$ if $Y_{n1},$ the number of exceedances of $v_n$ in a block of size $r_n,$ is replaced by $\widetilde{N}_{r_n}(v_n).$ The proof now follows from considering $\alpha =1$ and $T(j)=j{1\mathrm{I}}_{\{j\ge 1\}}$ in this theorem for $\widetilde{N}_{r_n}(v_n)$ and verifying that conditions (a), (c) and (d) of this theorem follow from the assumptions of Theorem 1, (2) and (3), (b) follows from Theorem 1 and (e) from the fact that

$$\begin{aligned} \lim _{n\rightarrow +\infty }\frac{k_n}{c_n} E\big [\widetilde{N}_{r_n}(v_n)\big ]=\lim _{n\rightarrow +\infty }\frac{k_n}{c_n} r_nP(X_\le v_n <X_2)=\nu . \end{aligned}$$

$\square$

1.3 Proof of Theorem 3

From Cramer-Wald’s Theorem we know that a necessary and sufficient condition for (2.5) to hold is that for any $a, b\in {\mathrm{I}R}$

$$\begin{aligned}&c_n^{-1/2}\left( a\displaystyle {\sum _{i=1}^{k_n}\left( \widetilde{N}_{r_n}^{(i)}(v_n)- E[\widetilde{N}_{r_n}^{(i)}(v_n)]\right) }+b \displaystyle {\sum _{i=1}^{k_n}\left( {1\mathrm{I}}_{\{\widetilde{N}_{r_n}^{(i)}(v_n)>0\}}- E[{1\mathrm{I}}_{\{\widetilde{N}_{r_n}^{(i)}(v_n)>0\}}]\right) }\right) \nonumber \\&\xrightarrow [n\rightarrow +\infty ]{d} \mathcal {N} (0,\ \nu (a^2\eta \sigma ^2+2ab+b^2\eta )). \end{aligned}$$

(5.4)

We shall therefore prove (5.4). For this, lets consider

$$\begin{aligned}&U_{r_n-l_n}^{(i)}(v_n)=\sum _{j=(i-1)r_n+1}^{ir_n-l_n}{1\mathrm{I}}_{\{X_j\le v_n<X_{j+1}\}},\quad \\&V_{l_n}^{(i)}(v_n)=\widetilde{N}_{r_n}^{(i)}(v_n)-U_{r_n-l_n}^{(i)}(v_n),\quad 1\le i\le k_n \end{aligned}$$

and note that

$$\begin{aligned}&a\sum _{i=1}^{k_n}\left( \widetilde{N}_{r_n}^{(i)}(v_n)- E[\widetilde{N}_{r_n}^{(i)}(v_n)]\right) +b \sum _{i=1}^{k_n}\left( {1\mathrm{I}}_{\{\widetilde{N}_{r_n}^{(i)}(v_n)>0\}}- E[{1\mathrm{I}}_{\{\widetilde{N}_{r_n}^{(i)}(v_n)>0\}}]\right) \nonumber \\&\quad =a\sum _{i=1}^{k_n}\left( U_{r_n-l_n}^{(i)}(v_n)-E[U_{r_n-l_n}^{(i)}(v_n)]\right) + a\sum _{i=1}^{k_n}\left( V_{l_n}^{(i)}(v_n)-E[V_{l_n}^{(i)}(v_n)]\right) \nonumber \\&\qquad +\, b\sum _{i=1}^{k_n}\left( {1\mathrm{I}}_{\{U_{r_n-l_n}^{(i)}(v_n)>0\}}- E[{1\mathrm{I}}_{\{U_{r_n-l_n}^{(i)}(v_n)>0\}}]\right) +\nonumber \\&\qquad +\, b\sum _{i=1}^{k_n}\left( {1\mathrm{I}}_{\{U_{r_n-l_n}^{(i)}(v_n)>0,\ U_{r_n-l_n}^{(i)}(v_n)=0\}}- E[{1\mathrm{I}}_{\{U_{r_n-l_n}^{(i)}(v_n)>0,\ U_{r_n-l_n}^{(i)}(v_n)=0\}}]\right) .\qquad \end{aligned}$$

(5.5)

Now, since (5.5) holds, to show (5.4) we have to prove the following

$$\begin{aligned}&c_n^{-1/2} \left( a\displaystyle \sum _{i=1}^{k_n}\left( U_{r_n-l_n}^{(i)}(v_n)-E[U_{r_n-l_n}^{(i)}(v_n)] \right) \right. \nonumber \\&\qquad \left. +\,b \sum _{i=1}^{k_n}\left( {1\mathrm{I}}_{\{U_{r_n-l_n}^{(i)}(v_n)>0\}}- E[{1\mathrm{I}}_{\{U_{r_n-l_n}^{(i)}(v_n)>0\}}]\right) \right) \nonumber \\&\xrightarrow [n\rightarrow +\infty ]{d} \mathcal {N} (0,\ \nu (a^2\eta \sigma ^2+2ab+b^2\eta )), \end{aligned}$$

(5.6)

$$\begin{aligned}&c_n^{-1/2} \sum _{i=1}^{k_n}(V_{l_n}^{(i)}(v_n)-E[V_{l_n}^{(i)}(v_n)])\xrightarrow [n\rightarrow +\infty ]{P}0,\end{aligned}$$

(5.7)

$$\begin{aligned}&c_n^{-1/2} \sum _{i=1}^{k_n}\left( {1\mathrm{I}}_{\{N_{r_n}^{(i)}>0,\ U_{r_n-l_n}^{(i)}(v_n)=0\}}-E[{1\mathrm{I}}_{\{N_{r_n}^{(i)}>0,\ U_{r_n-l_n}^{(i)}(v_n)=0\}}]\right) \xrightarrow [n\rightarrow +\infty ]{P}0.\qquad \quad \end{aligned}$$

(5.8)

Lets first prove (5.6). The summands in $\sum _{i=1}^{k_n}(aU_{r_n-l_n}^{(i)}(v_n)+b{1\mathrm{I}}_{\{U_{r_n-l_n}^{(i)}(v_n)>0\}})$ are functions of indicator variables that are at least $l_n$ time units apart from each other, therefore for each $t\in {\mathrm{I}R}$

$$\begin{aligned}&\left| E\left[ \exp \left( \mathrm{{i}}tc_n^{-1/2}\sum _{i=1}^{k_n}\left( aU_{r_n-l_n}^{(i)}(v_n)+ b{1\mathrm{I}}_{\{U_{r_n-l_n}^{(i)}(v_n)>0\}}\right) \right) \right] -\right. \end{aligned}$$

(5.9)

$$\begin{aligned}&-\left. \prod _{i=1}^{k_n}E\left[ \mathrm{{i}}tc_n^{-1/2}\left( aU_{r_n-l_n}^{(i)}(v_n)+ b{1\mathrm{I}}_{\{U_{r_n-l_n}^{(i)}(v_n)>0\}}\right) \right] \right| \le 16k_n\alpha _{n,l_n}, \end{aligned}$$

(5.10)

where, $\mathrm {i}$ is the imaginary unit, from repeatedly using a result in Volkonski and Rozanov (1959) and the triangular inequality. Now, since condition $\Delta (v_n)$ holds for $\mathbf{{X}}$ (5.10) tends to zero and so we can assume that the summands are i.i.d.. Therefore, in order to apply Lindberg’s Central Limit Theorem we need to verify that

$$\begin{aligned} \frac{k_n}{c_n} E\left[ \left( aU_{r_n-l_n}(v_n)+b{1\mathrm{I}}_{\{U_{r_n-l_n}(v_n)>0\}}\right) ^2\right] \xrightarrow [n\rightarrow +\infty ]{}\nu (a^2\eta \sigma ^2+2ab+b^2\eta ),\nonumber \\ \end{aligned}$$

(5.11)

since $\frac{k_n}{c_n}E^2[U_{r_n-l_n}(v_n)]\xrightarrow [n\rightarrow +\infty ]{}0,$ and Lindberg’s Condition

$$\begin{aligned}&\frac{k_n}{c_n}E\left[ \left( aU_{r_n-l_n}(v_n)+b{1\mathrm{I}}_{\{U_{r_n-l_n}(v_n)>0\}}\right) ^2 {1\mathrm{I}}_{\{(aU_{r_n-l_n}(v_n)+b{1\mathrm{I}}_{\{U_{r_n-l_n}(v_n)>0\}})^2>\epsilon c_n\}}\right] \nonumber \\&\quad \xrightarrow [n\rightarrow +\infty ]{}0,\quad \text {for all } \epsilon >0, \end{aligned}$$

(5.12)

with $U_{r_n-l_n}(v_n)=U_{r_n-l_n}^{(1)}(v_n).$

From the definition of $\widetilde{N}_{r_n}^{(i)}(v_n)$ and $V_{l_n}^{(i)}(v_n)$ and assumption (2) we have that

$$\begin{aligned} \frac{k_n}{c_n}E[V_{r_n}^2(v_n)]\le \left( \frac{r_n}{l_n}\right) ^{-1}\frac{k_n}{c_n}E[\widetilde{N}_{r_n}^2(v_n)]\xrightarrow [n\rightarrow +\infty ]{}0, \end{aligned}$$

(5.13)

with $V_{r_n}(v_n)=V_{r_n}^{(1)}(v_n).$ Now, by Cauchy-Schwarz’s inequality

$$\begin{aligned} \frac{k_n}{c_n}E[\widetilde{N}_{r_n}(v_n)V_{r_n}(v_n)]\le \sqrt{E[\widetilde{N}_{r_n}^2(v_n)]E[V_{r_n}^2(v_n)]}\xrightarrow [n\rightarrow +\infty ]{}0, \end{aligned}$$

thus

$$\begin{aligned} \frac{k_n}{c_n}E[U_{r_n-l_n}^2(v_n)]=\frac{k_n}{c_n} E[(\widetilde{N}_{r_n}(v_n)-V_{r_n}(v_n))^2]\xrightarrow [n\rightarrow +\infty ]{}\eta \nu \sigma ^2. \end{aligned}$$

(5.14)

On the other hand, since $k_n(r_n-l_n)\sim n,$ Theorem 1 implies that

$$\begin{aligned} \frac{k_n}{c_n}E\left[ {1\mathrm{I}}_{\{U_{r_n-l_n}(v_n)>0\}}\right] =\frac{k_n}{c_n} P(U_{r_n-l_n}(v_n)>0)\xrightarrow [n\rightarrow +\infty ]{}\eta \nu . \end{aligned}$$

(5.15)

Furthermore, by definition (2.3)

$$\begin{aligned}&\frac{k_n}{c_n}E\left[ U_{r_n-l_n}(v_n){1\mathrm{I}}_{\{U_{r_n-l_n}(v_n)>0\}}\right] \nonumber \\&\quad = \frac{k_n}{c_n}E\left[ U_{r_n-l_n}(v_n)\right] =\frac{k_n(r_n-l_n)}{c_n}P(X_1\le v_n<X_2)\xrightarrow [n\rightarrow +\infty ]{}\nu . \end{aligned}$$

(5.16)

(5.14)-(5.16) prove (5.11) and since Lindberg’s Condition follows immediately from assumption (1), (5.6) is proven.

Finally, since Theorem 1 of Hsing (1991) holds for $\widetilde{N}_{r_n}(v_n),$ it implies (5.7) and (5.8) because $\frac{k_n}{c_n}E[V_{l_n}^2(v_n)]\xrightarrow [n\rightarrow +\infty ]{}0$ by (5.13) and

$$\begin{aligned}&\frac{k_n}{c_n}E[{1\mathrm{I}}^2_{\{\widetilde{N}_{r_n}^{(i)}(v_n)>0,\ U_{r_n-l_n}^{(i)}(v_n)=0\}}]\\&\quad =\frac{k_n}{c_n}(P(\widetilde{N}_{r_n}^{(i)}(v_n)>0)- P(U_{r_n-l_n}^{(i)}(v_n))>0)\xrightarrow [n\rightarrow +\infty ]{}0 \end{aligned}$$

by Theorem 1 and (5.15). This concludes the proof. $\square$

1.4 Proof of Corollary 1

Since $\eta _n\xrightarrow [n\rightarrow +\infty ]{}\eta ,$ by Theorem 1, $\frac{k_n}{c_n}E[\widetilde{N}_{r_n}(v_n)]\xrightarrow [n\rightarrow +\infty ]{}\nu ,$ and $c_n^{-1}\sum _{i=1}^{k_n}(\widetilde{N}_{r_n}^{(i)}(v_n) -E[\widetilde{N}_{r_n}^{(i)}(v_n)])\xrightarrow [n\rightarrow +\infty ]{P}0,$ by Theorem 1 of Hsing (1991) which holds for $\widetilde{N}{r_n}^{(i)}(v_n),$ the result now follows from the fact that

$$\begin{aligned}&\sqrt{c_n}(\widehat{\eta }_n^B-\eta _n)=\frac{1}{c_n^{-1}\sum _{i=1}^{k_n} \widetilde{N}_{r_n}^{(i)}(v_n)}\left( c_n^{-1/2}\sum _{i=1}^{k_n} \left( {1\mathrm{I}}_{\{\widetilde{N}_{r_n}^{(i)}(v_n)>0\}}-E[{1\mathrm{I}}_{\{\widetilde{N}_{r_n}^{(i)}(v_n)>0\}}] \right) \right. -\\&\left. -\eta _n c_n^{-1/2}\sum _{i=1}^{k_n}\left( \widetilde{N}_{r_n}^{(i)}(v_n)- E[\widetilde{N}_{r_n}^{(i)}(v_n)]\right) \right) \end{aligned}$$

and Theorem 3. $\square$

1.5 Proof of Theorem 4

Since (5.2) holds, we only need to show that

$$\begin{aligned} \sum _{j=1}^{+\infty }(\widetilde{\pi }_n(j;\widehat{v}_n)-\widetilde{\pi }_n(j;v_n))\xrightarrow [n\rightarrow +\infty ]{P}0. \end{aligned}$$

(5.17)

Lets start by noting that for $v_n^{(\tau )}$ such that $P(X_1>v_n^{(\tau )})\sim c_n\tau /n$ as $n\rightarrow +\infty$ and $\epsilon >0$ we have

$$\begin{aligned}&\lim _{\epsilon \rightarrow 0}\lim _{n\rightarrow +\infty } \frac{k_n}{c_n}\left| E\left[ \widetilde{N}_{r_n}(v_n^{(\tau +\epsilon )}){1\mathrm{I}}_{\{\widetilde{N}_{r_n}(v_n^{(\tau +\epsilon )})>0\}} \right] -E\left[ \widetilde{N}_{r_n}(v_n^{(\tau )}){1\mathrm{I}}_{\{\widetilde{N}_{r_n}(v_n^{(\tau )})>0\}} \right] \right| \nonumber \\= & {} \lim _{\epsilon \rightarrow 0}\lim _{n\rightarrow +\infty }\frac{k_nr_n}{c_n}(P(X_1\le v_n^{(\tau +\epsilon )}<X_2)-P(X_1\le v_n^{(\tau )}<X_2))\nonumber \\= & {} \lim _{\epsilon \rightarrow 0}\lim _{n\rightarrow +\infty }\frac{k_nr_n}{c_n}(P(X_1> v_n^{(\tau +\epsilon )})-P(X_1> v_n^{(\tau )}))=0. \end{aligned}$$

(5.18)

(5.18) proves condition b) of Theorem 3 in Hsing (1991) which holds for $\widetilde{N}_{r_n}(v_n),$ where $T(j)=j{1\mathrm{I}}_{\{j\ge 1\}}.$ The other conditions have been verified in the proof of Theorem 2 as well as the conditions of Corollary 2.4 in Hsing (1991) for $\widetilde{N}_{r_n}(v_n).$ Therefore (5.17) holds, completing the proof. $\square$