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

Field Survey of 2018 Typhoon Jebi in Japan: Lessons for Disaster Risk Management, by Takabatake,T. et al., Geoscience, 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

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）

1959, September 26th Isewan Typhoon

* Large area innandated * Yodo River also damaged * Ship landed (like 2011)

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

Return Period Daily max Wind Speed [m/s] .5 1 2 5 10 20 50 100 200 15 20 25 30

Hamburg Hanover Bremerhaven Fehmarn Schleswig

Hanover Hamburg

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 ...

Hamburg Hanover Bremerhaven Fehmarn Schleswig

5 10 15 20 25 5 10 15 20 25 30 35

Wind Speed [m/s] at Hamburg Wind Speed [m/s] at Hanover

5 10 15 20 25 5 10 15 20 25 30 35

Wind Speed [m/s] at Hamburg Wind Speed [m/s] at Hanover

Return Period Daily max Wind Speed [m/s] .5 1 2 5 10 20 50 100 200 15 20 25 30

Rank(Hamburg) Rank(Hanover) 1 10 100 1000 10000 1 10 100 1000 10000

Hanover Hamburg

Corresponding relations

5 10 15 20 25 5 10 15 20 25 30 35

Wind Speed [m/s] at Hamburg Wind Speed [m/s] at Hanover

5 10 15 20 25 5 10 15 20 25 30 35

Wind Speed [m/s] at Hamburg Wind Speed [m/s] at Hanover

Return Period Daily max Wind Speed [m/s] .5 1 2 5 10 20 50 100 200 15 20 25 30

Rank(Hamburg) Rank(Hanover) 1 10 100 1000 10000 1 10 100 1000 10000

Hanover Hamburg

Return Period [Year] at Hamburg Return Period [Year] at Hanover . . 2 5 20 50 200 . . 2 5 20 50 200

Return Period for Joint Occurrence

Return Period Daily max Wind Speed [m/s] .5 1 2 5 10 20 50 100 200 15 20 25 30

Rank(Hamburg) Rank(Hanover) 1 10 100 1000 10000 1 10 100 1000 10000

Hanover Hamburg

1 10 100 1000 10000 0.0 0.2 0.4 0.6 0.8 1.0

Rank (decreasing Order), r Correlation Coef.

Hamburg & Hanover

x x x x x x x x x x x By formula By prod.−mom. By EM algo. r = 1281

r H(1) 100 1000 10000 0.5 1.0 1.5 2.0 1 10

• Thresh. Choice by the conventional method
• Thresh. Choice by the method of

correlation coeffjcient

• f occurrence numbers

Extremes by POT requires a threshold. (Peaks Over Threshold)

• Thresh. Choice should

be determined by the correlations, because the dependence problem can be solved by the properties of dependence. It is a logic.

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

• od . 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.

Fu = λ∗(yA ∧ uA, yB ∧ uB) − λ∗(yA, yB) λ∗(uA, uB)

D1 (uA, uB) (yA, yB) (uA, uB) (yA, yB) D2

¯ Fu(yA, yB) = 1 − Fu(yA, yB) = λ∗(yA, yB) λ∗(uA, uB)

case 1 case 2 Here let us give a glance to the probability distribution:

Fu(yA, yB) = λB(uB) − λB(yB) − {λAB(yA, uB) − λAB(yA, yB)} λ∗(uA, uB)

where we make the function λ∗ endlessly, ... One of the simplest ones is:

λ∗(yA, yB) =

• λ 1/α

A (yA) + λ 1/α B (yB)

α

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 orro o o orr r,

• y md ro ord 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. dependent <--- ---> independent Kasukabe to Tsukuba, Kamagatani, Sano

Hamburg Hanover 5 10 15 20 25 5 10 15 20 25 30 Hamburg Hanover 5 10 15 20 25 5 10 15 20 25 30

Counting the excess numbers is another way of evaluating extremes, ...

kA(uA) =

n

• i=1

1{YA(i) > uA}, kB(uB) =

n

• i=1

1{YB(i) > uB}

kA = 4 kB = 3 uB uA

Hamburg Hanover 5 10 15 20 25 5 10 15 20 25 30 Hamburg Hanover 5 10 15 20 25 5 10 15 20 25 30

k∗ = 5 kAB = 2 (uA, uB) uB uA

Inclusive occurrence number Joint occurrence number

k∗(uA, uB) =

n

• i=1

1{YA(i) > uA} ∨ 1{YB(i) > uB} kAB(uA, uB) = kA(uA) + kB(uB) − k∗(uA, uB)

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: The case of no occurrence gives the cumulative distribution (= non-exceedance probability) function: where the rate function is set to then Eq.(*) corresponds to a GEV (Generalized Extreme Value) distribution.

f(k) = λk

1

k! e−λ1 f(k = 0) = e−λ1 → λ1(y) =

• 1 + ξ y − µ1

σ1 −1/ξ G1(y) = exp

• −
• 1 + ξ y − µ1

σ1 −1/ξ G1(y) = e− λ1

• λ1 = λ1(y) = e− λ1(y)

Key'2’:''A'Poisson'distribu=on'into'an'extreme'value'd.' *'Univariate'case:'

p(kx = 0) = {λ1(x)}0 0! e−λ1(x) = exp ( − ✓ 1 + ξx x − µx,1 σx,1 ◆− 1/ξx)

No'occurrence'prob.''' '''''''''''''''''='cumula=ve'prob.'distribu=on'of'max. *'Bivariate'case:' Therefore'the'inclusive'occurrence'rate'becomes'' Important'in'the'theore=cal'treatment.

p(kx,1 = 0, ky,1 = 0) = e−λ∗,1(x,y) = F1(x, y)

= F1(x)

λ∗,1(x, y) = E(k∗) = E X

i

1 {Xi > x} ∨ 1 {Yi > y}

= Z 1 Z ∞

x1 t ∧ y1 1−t

drdH(t) r2

= Z 1 ✓ t x1 ∨ 1 − t y1 ◆ dH(t)

= Z 1 {tλ1(x) ∨ (1 − t)λ1(y)} dH(t)

A(ω) = λ∗,1(x, y) λ1(x) + λ1(y) = Z 1 {t(1 − ω) ∨ (1 − t)ω} dH(t)

ω = λ1(y) λ1(x) + λ1(y) = x1 x1 + x2

= Z 1 ⇢ r > x1 t ∧ y1 1 − t

• dV (r, t)

= Z 1 ⇢ x0

1

x1 ∨ y0

1

y1 > 1

• dV (x0

1, y0 1)

y0

1 = r(1 − t)

x0

1 = rt

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E%$579'E9@"-'.99-784"#$% X'8%'.94%#-"84$7'94@C'YC&

Z 1 dH(t) = 2

Z 1 tdH(t) = Z 1 (1 − t)dH(t) = 1

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.

Rank for Wind Speed in Hamburg Rank for Wind Speed in Hanover 40 1 10 100 1000 10000 20 1 10 100 1000 10000 Rank for Wind Speed in Hamburg Rank for Wind Speed in Hanover 8 1 100 1000 10000 1 10 100 1000 10000 4

25 15 5 25 + 15 = 40 5 + 3 = 8 5 + 15 = 20 1 + 3 = 4 5 3 1 where we know ranks in decreasing order indicate the occurrence numbers.

Correlation coef. of occurrence numbers is given by which is based on the bivariate Poisson distribution, and it will be estimated by sample, as for the common number of rank j.

• cf. c.c. by prod. mom. est.

ρxy = λxy

• λx λy

˜ ρxy = (kx − ¯ kx)(ky − ¯ ky) (kx − ¯ kx)2 (ky − ¯ ky)2

Rank(Hamburg) Rank(Hanover)

1 10 100 1000 10000 1 10 100 1000 10000 1 10 100 1000 10000 0.0 0.2 0.4 0.6 0.8 1.0 Rank (decreasing Order), j Correlation Coef. Hamburg & Hanover x x x x x x By formula By prod.−mom. r = 1281

ˆ ρxy = kxy j

Con=ngency'tables'for'excess'&'no'excess Perfect'dependent

↔ kxy = 0

k∗ = kx ∨ ky k∗ = kx + ky

Independent

ux \ uy Excess No Excess Total Excess kxy kx − kxy kx No Excess ky − kxy (n − kx − ky + kxy) (n − kx) Total ky (n − ky) (n)

ux \ uy Excess No Excess Total Excess 20 25 45 No Excess 19 (667) (686) Total 39 (692) (731)

ux \ uy Excess No Excess Total Excess 45 45 No Excess 39 (647) (686) Total 39 (692) (731) ux \ uy Excess No Excess Total Excess 39 6 45 No Excess (686) (686) Total 39 (692) (731)

Daily'max.' 'wind'speeds' for''2'years

r H(1) 1 10 100 1000 10000 0.5 1.0 1.5 2.0

Conventional method uses the identity equation: what this equation stands for?

• > The occurrrence numbers are the same in the following two regions.

(The red line is a contourline of 1/rank_A + 1/rank_B.)

H(1) = 2

However this method will be overestimated, as seen in this example where we can take r= 1281, while the C.C. is not stable. Actually the stable data is limited around 100.

Sample Pareto Margin for Hamburg Sample Pareto Margin for Hanover .0002 .001 .01 .1 1 .0002 .001 .01 .1 1

1 10 100 1000 10000 0.0 0.2 0.4 0.6 0.8 1.0 Rank (decreasing Order), j Correlation Coef. Hamburg & Hanover x x x x x x By formula By prod.−mom. r = 1281

The important thing is that demension reduction is possible by transforming the occurrence rate λ∗(x, y) into the Pickands dependence function A(t). (Bivariate extreme distribution is so simple that there are included the wide range of the mathematical functions for the distributions.)

λ∗(x, y) = E(k∗) = E

• i

1{Xi > x} ∨ 1{Yi > y} = · · · omitting the details of derivation · · · = 1 {ωλA(x)} ∨ {(1 − ω)λB(y)} dH(ω) A(t) = 1 {ω(1 − t)} ∨ {(1 − ω)t} dH(ω) t = λB(y) λA(x) + λB(y) = 1/λA(y) 1/λA(x) + 1/λB(y)

r d rr r and the radius (lengthwise) variable It is the pseudo polar coordinates.

r = 1/λA(x) + 1/λB(y)

t A(t) 0.0 0.5 1.0 0.5 1.0

independent p e r f e c t d e p e n d e n t p e r f e c t d e p e n d e n t

A(t) ≡ 1 A(t) = t A(t) = 1 − t

t Density 0.0 0.5 1.0 0.0 0.5 1.0 1.5 0.005 < r < 0.02

Sample Pareto Margin for Hamburg Sample Pareto Margin for Hanover .0002 .001 .01 .1 1 .0002 .001 .01 .1 1 r=0.02 r=0.01 r=0.005

Pickands dependence function shows well the dependency properties.

ti = pi/ri ri = pi + qi pi (= 1/rankA) qi (= 1/rankB)

pseudo-polar coordinate

˜ Ai = k∗ rankA + rankB

We analyze the spatial dependence of daily precipitations by using d4PDF.

<- Cor.

Coef.

<- Pickands Dependent Function <- Transverse Density <- Contours

• f Retrun

Period of joint

• ccurrences

Conclusions

What we show here today is actually not new in extreme value theory. But these facts have not been well known for hydrology and water-lerated engineerings (Coastal and Hydraulic Eng.). Parametric approach as well as non-parametric approach should be applied, because the wide range of parametric functions is possible to describe the joint (or inclusive) occurrence rate. It is so important to examine the joint occurrence rates (and the return period) for the accumulative risk. For the future works, the bivariate extreme analysis should be extended to the spatial modeling of the whole river basin and the comprehensive coastal zone.

Return Period [Year] at Hamburg Return Period [Year] at Hanover . . 2 5 20 50 200 . . 2 5 20 50 200