Applications of bivariate generalized Pareto distribution and the - PowerPoint PPT Presentation

Applications of bivariate generalized Pareto distribution and the threshold choice Toshikazu Kitano Nagoya Institute of Technology

One of the biggest disasters in Japan, 2018

Mori, N. & T. Yasuda, T. Arikawa, T. Kataoka, S. Nakajo, K. Suzuki, Y. Yamanaka: 2018 Typhoon Jebi Post-Event Survey of Coastal Damage in the Kansai Region, Japan. Coastal Engineering Journal Field Survey of 2018 Typhoon Jebi in Japan: Lessons for Disaster Risk Management, by Takabatake,T. et al., Geoscience, 2018.

台風２１号による高潮は， 第二室戸台風（昭和３６年）を越える規模（ほぼ同程度） 淀川での高潮の河川遡上 ，高潮による水位が堤防高を超過． 大阪府の３大水門の閉鎖による浸水回避 ． 淀川本川の３橋の防潮鉄扉（陸閘）の閉鎖（１９７９年以来）

* Large area innandated * Yodo River also damaged * Ship landed (like 2011) 1959, September 26th Isewan Typhoon

This year 2019, September 9th Tyhoon no.15 Faxai behaved violently, ...

Dependency of Extremes When spatially near communities are attacked at the same time , disaster will expand more . Support, Back-up & Recovery will become more diffjcult than ... 30 Hanover Daily max Wind Speed [m/s] 25 Schleswig Fehmarn Bremerhaven Hamburg 20 Hanover Hamburg 15 .5 1 2 5 10 20 50 100 200 Return Period

35 35 10000 Wind Speed [m/s] at Hanover Wind Speed [m/s] at Hanover 30 30 1000 25 25 Rank(Hanover) 20 20 100 15 15 10 10 10 5 5 1 0 0 0 5 10 15 20 25 0 5 10 15 20 25 1 10 100 1000 10000 Wind Speed [m/s] at Hamburg Wind Speed [m/s] at Hamburg Rank(Hamburg) 30 Hanover Daily max Wind Speed [m/s] 25 Schleswig Fehmarn Bremerhaven Hamburg 20 Hanover Hamburg 15 Corresponding relations .5 1 2 5 10 20 50 100 200 Return Period

35 35 10000 Wind Speed [m/s] at Hanover Wind Speed [m/s] at Hanover 30 30 1000 25 25 Rank(Hanover) 20 20 100 15 15 10 10 10 5 5 1 0 0 0 5 10 15 20 25 0 5 10 15 20 25 1 10 100 1000 10000 Wind Speed [m/s] at Hamburg Wind Speed [m/s] at Hamburg Rank(Hamburg) 30 Return Period for Joint Occurrence Hanover 200 Daily max Wind Speed [m/s] 25 Return Period [Year] at Hanover 50 20 20 5 2 Hamburg . 15 . . . 2 5 20 50 200 .5 1 2 5 10 20 50 100 200 Return Period [Year] at Hamburg Return Period

10000 Extremes by POT requires a threshold. (Peaks Over Threshold) 1000 Rank(Hanover) Thresh. Choice by the conventional method 100 Thresh. Choice should 2.0 be determined by the 1.5 correlations, because 10 H(1) the dependence problem 1.0 can be solved by the 1 properties of dependence. 0.5 1 10 100 1000 10000 It is a logic. Rank(Hamburg) 1 10 100 1000 10000 r 1.0 30 By formula x By prod.−mom . Hanover 0.8 Correlation Coef. x By EM algo. Daily max Wind Speed [m/s] x x 0.6 25 x x 0.4 x x x 0.2 x 20 x Hamburg & Hanover 0.0 r = 1281 Hamburg 1 10 100 1000 10000 15 Rank (decreasing Order), r Thresh. Choice by the method of correlation coeffjcient .5 1 2 5 10 20 50 100 200 of occurrence numbers Return Period

http://www.jamstec.go.jp/tougou/program/index.html

To evaluate the uncertainty requires the extreme value theory and the statistical techniques to the applications.

http://www.miroc-gcm.jp/~pub/d4PDF/index_en.html

The importance of bivariate extreme statistics: Overlap of several hazards: storm surge, high waves, river runoff and �ood��� ���. accumulates risk and its prediction will become troublesome. Simultaneous occurrences (or joint occurrences) at several sites (at least two important sites) also aggregate the loss by damage, and they will enlarge the loss more than the proportional one. The importance of statistical distribution of bivariate extremes is now increasing in disaster risk reduction plan, but the bivariate GP distribution has been not yet developed enough for those applications.

One of the reasons is the unclearness of mathematical understanding of the multivariate (bivariate) extremes for the practical engineers. Here let us give a glance F u = λ ∗ ( y A ∧ u A , y B ∧ u B ) − λ ∗ ( y A , y B ) to the probability distribution: λ ∗ ( u A , u B ) case 1 ( y A , y B ) F u ( y A , y B ) = 1 − F u ( y A , y B ) = λ ∗ ( y A , y B ) ¯ D 1 λ ∗ ( u A , u B ) ( u A , u B ) case 2 D 2 ( y A , y B ) F u ( y A , y B ) = λ B ( u B ) − λ B ( y B ) − { λ AB ( y A , u B ) − λ AB ( y A , y B ) } λ ∗ ( u A , u B ) ( u A , u B ) where we make the function λ ∗ endlessly, ... One of the simplest ones is: � α � λ 1 / α A ( y A ) + λ 1 / α λ ∗ ( y A , y B ) = B ( y B )

And therefore, though GP distribution requires a suitable threshold to extract the extremes of hazard magnitudes, the methods of threshold choice have not been enough discussed for bivariate extremes. This research focuses on the �orr�����o� �o�������� o� o���rr���� r���� , ��o�� ��������y �� ���m���d ��ro��� ��� o���r��d d��� o� wind velocities at two cities. And the numerous datasets of daily rainfall in d4PDF are also applied to nonparametric analysis of bivariate extremes to demonstrate the spacial change of depemdence of the pairwise points against the distances. Kasukabe to Tsukuba, Kamagatani, Sano dependent <--- ---> independent

Counting the excess numbers is another way of evaluating extremes, ... 30 30 k A = 4 k B = 3 25 25 u B 20 20 Hanover Hanover 15 15 10 10 5 5 u A 0 0 0 5 10 15 20 25 0 5 10 15 20 25 Hamburg Hamburg n n � � k A ( u A ) = 1 { Y A ( i ) > u A } , k B ( u B ) = 1 { Y B ( i ) > u B } i =1 i =1 Inclusive occurrence number Joint occurrence number 30 30 k AB = 2 k ∗ = 5 25 25 u B 20 20 ( u A , u B ) Hanover Hanover 15 15 10 10 5 5 u A 0 0 0 5 10 15 20 25 0 5 10 15 20 25 Hamburg Hamburg n � k ∗ ( u A , u B ) = 1 { Y A ( i ) > u A } ∨ 1 { Y B ( i ) > u B } k AB ( u A , u B ) = k A ( u A ) + k B ( u B ) − k ∗ ( u A , u B ) i =1

For one-component, it will be easier to understand the relation between the Poisson disitribution and the extreme variable. The uni-variate Poisson distribution is described in terms of the mean rate as follows: f ( k ) = λ k k ! e − λ 1 1 The case of no occurrence gives the cumulative distribution (= non-exceedance probability) function: λ 1 = λ 1 ( y ) = e − λ 1 ( y ) G 1 ( y ) = e − λ 1 � f ( k = 0) = e − λ 1 → � � − 1 / ξ � 1 + ξ y − µ 1 where the rate function is set to λ 1 ( y ) = σ 1 then Eq.(*) corresponds to a GEV (Generalized Extreme Value) distribution. � � − 1 / ξ � � 1 + ξ y − µ 1 G 1 ( y ) = exp − σ 1

Key'2’:''A'Poisson'distribu=on'into'an'extreme'value'd.' *'Univariate'case:' � ( ◆ − 1 / ξ x ) p ( k x = 0) = { λ 1 ( x ) } 0 ✓ x − µ x, 1 e − λ 1 ( x ) = exp 1 + ξ x − 0! σ x, 1 = F 1 ( x ) No'occurrence'prob.''' '''''''''''''''''='cumula=ve'prob.'distribu=on'of'max. � *'Bivariate'case:' � p ( k x, 1 = 0 , k y, 1 = 0) = e − λ ∗ , 1 ( x,y ) = F 1 ( x, y ) Therefore'the'inclusive'occurrence'rate'becomes'' Important'in'the'theore=cal'treatment. �

X λ ∗ , 1 ( x, y ) = E( k ∗ ) = E 1 { X i > x } ∨ 1 { Y i > y } i ⇢ x 0 E%$579'E9@"-'.99-784"#$% � ∨ y 0 � Z 1 1 dV ( x 0 1 , y 0 = 1 > 1 1 ) x 0 1 = rt x 1 y 1 y 0 1 = r (1 − t ) ⇢ � Z r > x 1 y 1 = 1 dV ( r, t ) ∧ 1 − t t X'8%'.94%#-"84$7'94@C'YC& � Z 1 Z ∞ drdH ( t ) = Z 1 r 2 x 1 y 1 0 dH ( t ) = 2 t ∧ 1 − t ✓ t Z 1 0 Z 1 Z 1 ∨ 1 − t ◆ ^84"@@C'u � = dH ( t ) tdH ( t ) = (1 − t ) dH ( t ) = 1 x 1 y 1 0 0 0 Z 1 = { t λ 1 ( x ) ∨ (1 − t ) λ 1 ( y ) } dH ( t ) 647':9-$0'g � 0 i8.A"47%'7$E$47$4#'>54.=94 � Z 1 λ ∗ , 1 ( x, y ) A ( ω ) = λ 1 ( x ) + λ 1 ( y ) = { t (1 − ω ) ∨ (1 − t ) ω } dH ( t ) 0 λ 1 ( y ) x 1 ω = λ 1 ( x ) + λ 1 ( y ) = x 1 + x 2

Essential for bivariate extremes: Homogeneity (of order - 1) whose property shows the proportionality & similarity of the occurrence rate. And it will be checked by using the sample data. 10000 10000 25 5 Rank for Wind Speed in Hanover Rank for Wind Speed in Hanover 1000 1000 100 100 20 10 10 4 15 5 3 1 1 1 1 10 40 100 1000 10000 1 8 100 1000 10000 Rank for Wind Speed in Hamburg Rank for Wind Speed in Hamburg 25 + 15 = 40 5 + 3 = 8 5 + 15 = 20 1 + 3 = 4 where we know ranks in decreasing order indicate the occurrence numbers.

λ xy ρ xy = Correlation coef. of occurrence numbers is given by � λ x λ y which is based on the bivariate Poisson distribution, and it will be estimated by sample, as ρ xy = k xy ˆ j for the common number of rank j. � ( k x − ¯ k x )( k y − ¯ k y ) cf. c.c. by prod. mom. est. ρ xy = ˜ �� ( k x − ¯ k x ) 2 � ( k y − ¯ k y ) 2 10000 1.0 By formula x By prod.−mom . 0.8 Correlation Coef. Rank(Hanover) 1000 x 0.6 x 100 0.4 x x 0.2 x 10 Hamburg & Hanover 0.0 r = 1281 1 1 10 100 1000 10000 1 10 100 1000 10000 Rank (decreasing Order), j Rank(Hamburg)