A stochastic methodology for risk assessment of a large earthquake when a long time has elapsed

Abstract

We propose a stochastic methodology for risk assessment of a large earthquake when a long time has elapsed from the last large seismic event. We state an approximate probability distribution for the occurrence time of the next large earthquake, by knowing that the last large seismic event occurred a long time ago. We prove that, under reasonable conditions, such a distribution is exponential with a rate depending on the asymptotic slope of the cumulative intensity function corresponding to a nonhomogeneous Poisson process. As it is not possible to obtain an empirical cumulative distribution function of the waiting time for the next large earthquake, an estimator of its cumulative distribution function based on existing data is derived. We conduct a simulation study for detecting scenario in which the proposed methodology would perform well. Finally, a real-world data analysis is carried out to illustrate its potential applications, including a homogeneity test for the times between earthquakes.

Appendix

Proof of Proposition 2.1

Let L and $L^\lambda$ be an HPP and an NHPP, respectively, where L has rate equal to one and $L^\lambda$ has intensity $\lambda$. Let $\tau _k$ and $\tau _k^\lambda$ be the kth jump of L and $L^\lambda$, respectively. Consequently, $T_k$ and $\tau _k^\lambda$ are equal in distribution and since $L^\lambda _t$ has the same distribution that $L_{\Lambda (t)}$, we have

$$\begin{aligned} {\mathbb{P}}(T_k\le t) & = {\mathbb{P}}(\tau ^\lambda _k\le t) = {\mathbb{P}}(L^\lambda _t\ge k) = {\mathbb{P}}(L_{\Lambda (t)}\ge k) = {\mathbb{P}}(\tau _k\le \Lambda (t))\\ & = \int _{0}^{\Lambda (t)}\frac{1}{(k-1)!}u^{k-1}\exp (-u){\rm d}u\\ & = \int _{0}^{t}\frac{\lambda (r)}{(k-1)!}\Lambda (r)^{k-1}\exp (-\Lambda (r)){\rm d}r,\quad ({\text{by making }} u=\Lambda (r)). \end{aligned}$$

Therefore, the PDF of $T_k$ is given by (2.4). $\square$

Proof of Theorem 3.1

Let F be the CDF of T. Then, by the L’Hôpital rule and the fundamental theorem of calculus, we have

$$\begin{aligned} G(h) & = \lim _{t\rightarrow \infty } {\mathbb{P}}(t<T\le t+h|T>t) \\ & = 1 - \lim _{t\rightarrow \infty }\frac{1-F(t+h)}{1-F(t)} = 1 - \lim _{t\rightarrow \infty }\frac{F'(t+h)}{F'(t)} = 1 - \lim _{t\rightarrow \infty }\frac{f(t+h)}{f(t)}. \end{aligned}$$

Consequently,

$$\begin{aligned} \frac{G(h+\Delta h)-G(h)}{\Delta h} & = \frac{1}{\Delta h}\left( \lim _{t\rightarrow \infty }\frac{(f(t+h)-f(t+h+\Delta h))}{f(t+h)} \frac{f(t+h)}{f(t)}\right) \nonumber \\ & = \frac{1}{\Delta h}\left( 1-\lim _{t\rightarrow \infty }\frac{f(t+h+\Delta h)}{f(t+h)}\right) \lim _{t\rightarrow \infty }\frac{f(t+h)}{f(t)}.\nonumber \\ & = \frac{G(\Delta h)(1-G(h))}{\Delta h}. \end{aligned}$$

(8.1)

Thus, by taking limit as $\Delta h \rightarrow 0$ in (8.1), we have that ${{\text{d}}}G(h)/{\text{d}}h = G'(h)=G'(0)(1-G(h))$. This differential equation has a unique solution, for $G'(0)=m$, which is given by $G(h)=1-\exp (-m h)$. Therefore, the proof is complete. $\square$

Proof of Theorem 3.2

From the L’Hôpital rule, we have

$$\lim _{t\rightarrow \infty }\frac{\Lambda (t+h)}{\Lambda (t)}=\lim _{t\rightarrow \infty }\frac{\lambda (t+h)}{\lambda (t)}=1.$$

In addition, from the mean value theorem, there exists $\xi (t)$ between t and $t+h$, with $t >0$ and $h\ge 0$, such that $\Lambda (t+h)-\Lambda (t)=\lambda (\xi (t))h$. Consequently,

$$\frac{f_k(t+h)}{f_k(t)}=\exp (-\lambda (\xi (t))h)\left( \frac{\Lambda (t+h)}{\Lambda (t)}\right) ^{k-1}\frac{\lambda (t+h)}{\lambda (t)}.$$

Thus, because

$$\lim _{t\rightarrow \infty }\frac{f_k(t+h)}{f_k(t)}=\lim _{t\rightarrow \infty }\exp (-\lambda (\xi (t))h)=\exp (-m h),$$

the proof follows from (i) in Theorem 3.1. $\square$

Proof of Theorem 4.1

For each $\varepsilon >0$, let $t_0>0$, such that $|\lambda (s)-m|<\varepsilon$, whenever $s\ge t_0$. Consequently, for each $t>t_0$, we have

$$\begin{aligned} \left| \frac{1}{t}\int _0^t\lambda (s) {\rm d}s -m\right| & = \left| \frac{1}{t}\int _0^t(\lambda (s)-m) {\rm d}s\right| \\ & \le \frac{1}{t}\int _0^t|\lambda (s)-m| {\rm d}s \\ & \le \frac{1}{t}\int _0^{t_0}|\lambda (s)-m| {\rm d}s + \frac{1}{t}\int _{t_0}^t|\lambda (s)-m| {\rm d}s\\ & < \frac{1}{t}\int _0^{t_0}|\lambda (s)-m| {\rm d}s +\varepsilon . \end{aligned}$$

By taking limit as $t\rightarrow \infty$, we obtain

$$\begin{aligned} \limsup _{t\rightarrow \infty }\left| \frac{1}{t}\int _0^t\lambda (s) {\rm d}s -m\right| \le \varepsilon . \end{aligned}$$

As $\varepsilon >0$ is arbitrary, we have $\lim _{t\rightarrow \infty }{\Lambda (t)}/{t} = m$, which concludes the proof. $\square$

Proof of Theorem 4.2

Let $M=\{M_t;t\ge 0\}$ be the martingale defined by $M_t=N_t-\Lambda (t)$. As

$$\frac{N_t-N_{\tau ^*}}{t-\tau ^*}=\frac{\Lambda (t)-\Lambda (\tau ^*)}{t-\tau ^*}\left( \frac{M_t-M_{\tau ^*}}{\Lambda (t)-\Lambda (\tau ^*)}+1\right) ,$$

and Theorem 8.2.17 in Dacunha-Castelle and Duflo (1986) implies $\lim _{t\rightarrow \infty }(M_t-M_{\tau ^*})/(\Lambda (t)-\Lambda (\tau ^*))=0$, ${\mathbb{P}}$-a.s., from Theorem 4.1, we have $\lim _{n\rightarrow \infty }{(N_t-N_{\tau ^*})}/{(t-\tau ^*)}=m$, ${\mathbb{P}}$-a.s.

Suppose that $\lim _{t\rightarrow \infty }\sqrt{t}({\Lambda (t)}/{t}-m)=0$ and let L be an HPP with rate equal to one. For each $t\ge \tau ^*$, we have

$$\sqrt{t-\tau ^*}\left( \frac{L_{\Lambda (t)}-L_{\Lambda (\tau ^*) }}{t-\tau ^*}-m\right) = \left( \frac{\Lambda (t)}{t-\tau ^*}\right) ^{1/2} \left( \frac{L_{\Lambda (t)}-\Lambda (t)}{\sqrt{\Lambda (t)}}\right) +\sqrt{t-\tau ^*}\left( \frac{\Lambda (t)-L_{\Lambda (\tau ^*) }}{t-\tau ^*}-m\right) .$$

Consequently, to prove that

$$\sqrt{t-\tau ^*}\left( \frac{L_{\Lambda (t)}-L_{\Lambda (\tau ^*) }}{t-\tau ^*}-m\right) \mathop {\rightarrow } \limits ^{\mathcal {D}} {\rm N}(0,m), \quad {\text{ as }} \quad t \rightarrow \infty ,$$

(8.2)

and since $\lim _{t\rightarrow \infty } L_{\Lambda (\tau ^*)}/\sqrt{t-\tau ^*}=0$, we need to prove

$$\left( \frac{\Lambda (t)}{t-\tau ^*}\right) ^{1/2} \left( \frac{L_{\Lambda (t)}-\Lambda (t)}{\sqrt{\Lambda (t)}}\right) \mathop {\rightarrow } \limits ^{\mathcal {D}} {\rm N}(0,m),\quad {\text{ as }}\quad t \rightarrow \infty , \quad {\text{and}}$$

(8.3)

$$\lim _{t\rightarrow \infty }\sqrt{t-\tau ^*}\left( \frac{\Lambda (t)}{t-\tau ^*}-m\right) =0.$$

(8.4)

As $(L_t-t)/\sqrt{t}\mathop {\rightarrow } \limits ^{\mathcal {D}} {\rm N}(0,1)$ and $\lim _{t\rightarrow \infty }\Lambda (t)/(t-\tau ^*)=m$, condition (8.3) follows. From the intermediate value theorem, for each $t>0$, there exists $\xi (t)$ between $t-\tau ^*$ and t such that $\Lambda (t) = \Lambda (t-\tau ^*)+\lambda (\xi (t))\tau ^*$. This fact implies that

$$\sqrt{t-\tau ^*}\left( \frac{\Lambda (t)}{t-\tau ^*}-m\right) =\sqrt{t-\tau ^*}\left( \frac{\Lambda (t-\tau ^*)}{t-\tau ^*}-m\right) + \frac{\lambda (\xi (t))\tau ^*}{\sqrt{t-\tau ^*}}.$$

Because $\lim _{t\rightarrow \infty }\sqrt{t-\tau ^*}({\Lambda (t-\tau ^*)}/{(t-\tau ^*)}-m)=0$ is assumed and $\lambda$ is bounded, condition (8.4) holds. Hence, we have proven condition (8.2), but due to $N_t-N_{\tau ^*}$ and $L_{\Lambda (t)}-L_{\Lambda (\tau ^*)}$ have the same distribution, for each $t>0$, we have

$$\sqrt{t-\tau ^*}\left( \frac{N_t-N_{\tau ^*}}{t-\tau ^*}-m\right) \mathop {\rightarrow }\limits ^{\mathcal {D}}\,{\rm N}(0,m),\quad {\text{ as }}\quad t \rightarrow \infty ,$$

which completes the proof. $\square$

Proof of Theorem 5.1

From (i) in Theorem 4.2, for each $h>0,$ $\widehat{G}_\tau (h)\mathrel {\mathop {\longrightarrow }\limits ^{}_{n\rightarrow \infty }} G(h),$ ${\mathbb{P}}$-a.s. Hence, it follows from the Pólya Lemma (Roussas, 1997) that, as $\tau \rightarrow \infty$, $\{\widehat{G}_\tau ; \tau >0\}$ converges, ${\mathbb{P}}$-a.s., uniformly to the CDF G, which concludes the proof. $\square$

Proof of Theorem 5.2

From the intermediate value theorem, there exists $\xi _\tau$ between $\widehat{m}_\tau$ and m such that $\widehat{G}_\tau (h)-G(h)=h\exp (-\xi _\tau h)(m-\widehat{m}_\tau )$. Then,

$$\sup _{h\ge 0}\left|\sqrt{\tau }(\widehat{G}_\tau (h)-G(h))\right|=\frac{\exp (-1)}{\xi _\tau }\left|\sqrt{\tau }(\widehat{m}_\tau -m)\right|.$$

However, from (i) and (ii) of Theorem 4.2, $\lim _{\tau \rightarrow \infty }\xi _\tau =m$, ${\mathbb{P}}$-a.s., and $\sqrt{\tau }(\widehat{m}_\tau -m)\mathop {\rightarrow }\limits ^{\mathcal {D}}\,{\rm N}(0,m)$, respectively. Therefore, the proof follows from the Slutsky theorem. $\square$