Hydrological statistics and extremes

8/2/2019 Hydrological statistics and extremes

1/16

Hydrological statisticsHydrological statisticsand extremesand extremes

Marc F.P.Marc F.P. BierkensBierkens

Professor of HydrologyProfessor of Hydrology

Faculty of GeosciencesFaculty of Geosciences

Stochastic Hydrology

Hydrological statisticsHydrological statistics

Mostly concernes with the statistical analysisMostly concernes with the statistical analysis

of hydrological time series in relation toof hydrological time series in relation to

extremesextremes, i.e. floods and droughts., i.e. floods and droughts.


2/16

Example: Rhine at LobithExample: Rhine at Lobith

Daily averaged discharge (mDaily averaged discharge (m33/s) 1902-2002/s) 1902-2002

Example: Rhine at LobithExample: Rhine at Lobith

Flow duration curve of Rhine dischargeFlow duration curve of Rhine discharge


3/16

Extreme eventsExtreme events

Want to know:Want to know:

What is the probability distribution that a flood of given size occurs?What is the probability distribution that a flood of given size occurs?

What is the size of a flood that belongs to a given design frequency?What is the size of a flood that belongs to a given design frequency?

First question that must be answered is:First question that must be answered is:

What constitutes a flood?

Two methods:Two methods:

1.1. The largest discharge per year (Maximum values)The largest discharge per year (Maximum values)

2.2. All discharge values above a certain threshold (Peak over ThresholdAll discharge values above a certain threshold (Peak over Threshold

(POT) data or partial duration data)(POT) data or partial duration data)

Maximum values of Rhine at LobithMaximum values of Rhine at Lobith

Maximum daily averaged discharge (mMaximum daily averaged discharge (m33/s) for each year in 1902-2002/s) for each year in 1902-2002

0

2000

4000

6000

8000

10000

12000

14000

1902 1912 1922 1932 1942 1952 1962 1972 1982 1992 2002

Year

Qmax

(m3/s)


4/16

Assumptions about maximum valuesAssumptions about maximum values

1.1. The maximum values are realisations of independentThe maximum values are realisations of independent

random variables.random variables.

2.2. There is no trend in time.There is no trend in time.

3.3. The maximum values are identically distributed.The maximum values are identically distributed.

In this case a single probability distribution can beIn this case a single probability distribution can beassumed for the maximum values.assumed for the maximum values.

Probability and recurrence timesProbability and recurrence times

)Pr()( yYyF =Cumulative probability distribution:Cumulative probability distribution:

)Pr()( yYyP =Exceedence probability:Exceedence probability:

)(11

)(1)(

yFyPyT

==Return period or recurrence time:Return period or recurrence time:

Recurrence time: theRecurrence time: the averageaverage number years between two consecutivenumber years between two consecutive

flood events of a given sizeflood events of a given size

Note: the actual number of years between flood events of a given size isNote: the actual number of years between flood events of a given size is

itself random.itself random.


5/16

Recurrence times from dataRecurrence times from dataAnalysis of maximum values of Rhine discharge at Lobith for recurrence time

Y Rank F(y) P(y) T(y)

2790 1 0.00971 0.99029 1.0098

2800 2 0.01942 0.98058 1.0198

2905 3 0.02913 0.97087 1.0300

3061 4 0.03883 0.96117 1.0404

3220 5 0.04854 0.95146 1.0510

3444 6 0.05825 0.94175 1.0619

3459 7 0.06796 0.93204 1.0729

. . . . .

. . . . .9140 90 0.87379 0.12621 7.9231

9300 91 0.88350 0.11650 8.5833

9372 92 0.89320 0.10680 9.3636

9413 93 0.90291 0.09709 10.3000

9510 94 0.91262 0.08738 11.4444

9707 95 0.92233 0.07767 12.8750

9785 96 0.93204 0.06796 14.7143

9850 97 0.94175 0.05825 17.1667

10274 98 0.95146 0.04854 20.600011100 99 0.96117 0.03883 25.7500

11365 100 0.97087 0.02913 34.3333

11931 101 0.98058 0.01942 51.5000

12280 102 0.99029 0.00971 103.0000

)(1

1)(

)(1)(

1)(

yFyT

yFyP

n

iyF

=

=

+=

Large recurrence timesLarge recurrence times

From a series ofFrom a series ofnn maximum values: largestmaximum values: largest

recurrence time to be assessed:recurrence time to be assessed: nn+1 years!+1 years!

For design purposes:For design purposes:yyTT for largefor large TTis necessary!is necessary!

To this end:To this end:

1.1. Fit a probability distribution (which one?)Fit a probability distribution (which one?)

2.2. Use it to extrapolate to large values ofUse it to extrapolate to large values ofyyTT(maximum values(maximum valuesyy withwithFF((yy) close to 1.) close to 1.


6/16

The Gumbel distributionThe Gumbel distributionIt can be proven (see section 4.2.2. Syllabus) that maximum values ofIt can be proven (see section 4.2.2. Syllabus) that maximum values ofthe following distributions follow a Gumbel distribution:the following distributions follow a Gumbel distribution:

)exp()( )()( aybaybY ebeyf =

)exp()( )( aybY ezF=

-- ExponentialExponential

-- Gaussian

- logGaussian

- Gamma

- Logistic

- Gumbel

pdf:pdf:

cpdf:cpdf:

Fitting the Gumbel distributionFitting the Gumbel distribution

Types of methods:Types of methods:

graphical using Gumbel paper (linear regression)graphical using Gumbel paper (linear regression)

method of momentsmethod of moments

maximum likelihood estimationmaximum likelihood estimation


7/16

Gumbel paperGumbel paper

]})(

1)(ln[ln{

1

)]}(ln[ln{1

)()]}(ln[ln{

)]ln[(exp()](ln[)()(

yT

yT

bay

yFb

ay

aybyF

eeyF

T

YT

Y

aybayb

Y

=

=

=

==

Taking the double logarithm of the cpdf:Taking the double logarithm of the cpdf:

Plot maximumPlot maximumyyiivsvs

1

]}1

ln[ln{

+

n

i

Plot maximumPlot maximumyyii vsvs TT(y(yii) on special Gumbel) on special Gumbel

(double logarithmic) paper(double logarithmic) paper


1. Plot maximum1. Plot maximumyyii vsvs ]}1

ln[ln{+

n

i

or, plot maximumor, plot maximumyyii vsvs TT(y(yii) on special Gumbel) on special Gumbel

(double logarithmic) paper(double logarithmic) paper

2. Fit a straight line by eye or by regression2. Fit a straight line by eye or by regression

to determinato determina aa andand bb..


8/16


0.0005877,5604 == ba

Rhine data setRhine data set

Method of momentsMethod of moments

Mean and variance of a Gumbel variateMean and variance of a Gumbel variate YY::

ba

Y

5772.0+= 2

22

6bY

=

1) Estimate mean1) Estimate mean mmYY and varianceand variance2) Equate these to the above expressions2) Equate these to the above expressions

3) Solve for3) Solve foraa andand bb::

2

26

1

Ys

b=

2Ys

bma Y

5772.0 =


9/16

Method of momentsMethod of moments

0

2000

4000

6000

8000

10000

12000

14000

16000

18000

20000

1 10 100 1000 10000

Recurrence time (Years)

Qmax

(m3/s)

Rhine data setRhine data set

0.00061674,5621 == ba

Coffee breakCoffee break


10/16

Estimation of theEstimation of the TT-year event-year event

))1

ln(ln(

1

T

T

bayT

=

In case of the Rhine data set:In case of the Rhine data set:

MoM parameters:MoM parameters:

with parameters from regression:with parameters from regression:

./m17182))9992.0ln(ln(162156213

1250 sy ==

17736 m3/s

Estimation of confidence limitsEstimation of confidence limits

In case of regression:In case of regression:

)()( 9595 TstyyTsty YTTYT +

tt9595 : the 95-point of the students: the 95-point of the students tt-distribution-distribution

and the standard error of prediction estimated as:and the standard error of prediction estimated as:)( TsY

=

++=N

i

ii

iY

yyNxTx

xTx

NTs

1

2

2

22 )(

1

1

))((

))((11)(

withwith)]/)1ln((ln[)( iii TTTx =


11/16

Estimation of confidence limitsEstimation of confidence limits

In case of method of moments:In case of method of moments:

)(raV96.1)(raV96.1 TTTTT yyyyy +

withwith )]/)1ln((ln[)( TTTx =

Nb

xxyT 2

2

)61.052.011.1()(raV

++

Assuming a Gaussian estimation error:Assuming a Gaussian estimation error:

The number T-year events in a given periodThe number T-year events in a given period

TheThe expectedexpectednumber ofnumber of TT-year events in-year events inNNyears:years:

TNNp /=

TheThe actualactualnumbernumbernn ofof TT-year events in-year events in NN years is a randomyears is a random

variable obeying a binomial distribution:variable obeying a binomial distribution:

nNn

T ppn

NNyn

= )1(years)ineventsPr(

Examples:Examples: 0914.0)01.01(01.01

10years)10inevents1Pr( 9100 =

= y

Pr(one or more flood events occur in 10 years) = 1-Pr(no events) = 0956.099.01 10 = .


12/16

The time until the nextThe time until the next TT-year event-year event

TheThe expectedexpectednumber of yearsnumber of years nn until the nextuntil the next TT-year event:-year event: T

TheThe actualactualnumber of yearsnumber of years nn until the nextuntil the next TT-year event is a-year event is a

random variable obeying a geometric distribution (withrandom variable obeying a geometric distribution (withpp=1/=1/TT):):

ppym mT1)1()eventuntilyearsPr( =

Other extreme value distributionsOther extreme value distributionsGeneralized extreme value distributionsGeneralized extreme value distributions

Type 1Type 1

(Gumbel)(Gumbel)

Type IIIType III

(Weibull)(Weibull)

Type IIType II


13/16

Other extreme value distributionsOther extreme value distributions

Other maximum value distributions used for maximumOther maximum value distributions used for maximumvalues: log-normal, log-Pearson-type III.values: log-normal, log-Pearson-type III.

0

2000

4000

6000

8000

10000

12000

14000

16000

18000

20000

1 10 100 1000 10000

Recurrence time (Years)

Qmax

(m3/s)

Lognormal

Gumbel

Minimum values (e.g. low flows)Minimum values (e.g. low flows)

Take -Take -ZZor 1/or 1/ZZ

or use Weibull distribution onor use Weibull distribution onZZminmin


14/16

Testing the assumptionsTesting the assumptions

=

=+

=

n

i

i

n

i

ii

YY

YY

Q

1

2

1

1

2

1

)(

)(

Independence: Von Neumans QIndependence: Von Neumans Q

Lower critical area.

If larger than critical value: no evidence that data are dependent!

Rhine data set: Q=2.071 ; upper critical value at =0.05: 1.618 -> noevidence for dependence


Trends:Trends: Mann-Kendall testMann-Kendall test

=

=

=n

i

i

j

ji YYT2

1

1

)sgn(

)]52)(1(/[18' += nnnTT

ForFornn>40>40 TThas standard Gaussian distribution with two sided criticalhas standard Gaussian distribution with two sided critical

area (i.e. significant at 95% (area (i.e. significant at 95% (=0.05) accuracy trend is=0.05) accuracy trend is TT< -1.96 or< -1.96 or

TT> 1.96); otherwise no evidence of a trend.> 1.96); otherwise no evidence of a trend.

Rhine data set:Rhine data set: TT= 1.992 -> significant trend at 95% accuracy.= 1.992 -> significant trend at 95% accuracy.


15/16

Yearly maxima of average dayly runoff of the Rhine at Lobith

y = 12.682x - 18205

R2

= 0.0326

0

2000

4000

6000

8000

10000

12000

14000

1900 1920 1940 1960 1980 2000

Year

Qmax

(m3/s)


Testing for some distribution:Testing for some distribution: 22 testtest

1.1. Define aDefine a mm classes (as in a histogram) and assign theclasses (as in a histogram) and assign the nn data valuesdata values

2.2. Count the number of data falling in each each classCount the number of data falling in each each class ii:: nnii

3.3. Fit the proposed distribution functionFit the proposed distribution functionFF((YY) to the data) to the data

4.4. Calculate the expected number of data falling into each classCalculate the expected number of data falling into each class ii as:as:

5.5. The following test statistic is calculated:The following test statistic is calculated:

( ))()( lowup yFyFnn YYei =

=

=

m

ie

i

e

ii

n

nn

1

22 )(


16/16


Testing for some distribution:Testing for some distribution: 22 testtest

XX22 follows a chi-squared distribution withfollows a chi-squared distribution with mm-1 degress of freedom:-1 degress of freedom:

There is an upper critical area for the 0-hypothesis that the data followThere is an upper critical area for the 0-hypothesis that the data follow

the proposed distribution.the proposed distribution.

Rhine data and proposed Gumbel distribution and 20 classes:Rhine data and proposed Gumbel distribution and 20 classes:

Rhine data and proposed lognormal distribution and 20 classes:Rhine data and proposed lognormal distribution and 20 classes:

2

1m

70.232 =

36.142 =

Lower boundary critical area forLower boundary critical area formm-1 = 19 degrees of freedom-1 = 19 degrees of freedom

andand =0.05: 30.144 -> both distribution cannot be discarded!=0.05: 30.144 -> both distribution cannot be discarded!

Documents

Hydrological statistics and extremes