Hydrological statistics and extremes
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Transcript of Hydrological statistics and extremes
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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.
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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
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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)
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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.
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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.
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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
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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
Gumbel paperGumbel 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..
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Gumbel paperGumbel paper
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 =
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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
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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 =
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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 = .
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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
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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
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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
Testing the assumptionsTesting the assumptions
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.
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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 the assumptionsTesting the assumptions
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 )(
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Testing the assumptionsTesting the assumptions
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!