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Extreme Value Type - Stochastic Hydrology - Lecture Notes, Study notes of Mathematical Statistics

The main points i the stochastic hydrology are listed below:Extreme Value Type, Gumbel’s Extreme Value Distribution, Maximum Values, Double Exponential Distribution, Annual Peak Flood, Weibull Distribution, Mean and Variance, Parent Distribution, Sample Moment Values

Typology: Study notes

2012/2013

Uploaded on 04/20/2013

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Extreme Value Type-I Distribution
(Gumbelʼs Extreme Value Distribution)%
( ) ( )
{ }
( ) exp exp
;;0
fx x x
x
βα βα α
βα
⎡⎤
=−−
⎣⎦
−∞< < −∞< <>
mm
ʻʼ for maximum values and ʻ+ʼ for minimum values
Y = (X – β)/ α transformation
pdf
CDF –
[ ]
{ }
() exp expfy y y=mm
%<%y%<%%%
( )
{ }
( )
{ }
( ) exp exp
1 exp exp
Fy y
y
=
=
(maximum)
%
(minimum)
%
ˆ;
1.283
σ
α
=
ˆ0.45
0.45
βµσ
µσ
=
=+
(maximum)
%
(minimum)
%
(Double Exponential
Distribution)
%
Docsity.com
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pf4
pf5
pf8
pf9
pfa
pfd
pfe
pff
pf12
pf13
pf14
pf15
pf16
pf17
pf18
pf19
pf1a
pf1b
pf1c
pf1d
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Extreme Value Type-I Distribution

(Gumbel s Extreme Value Distribution)

{ }

( ) exp exp

f x x x

x

m m

  • for maximum values and + for minimum values

Y = (X – β)/ α → transformation

  • pdf
  • CDF –

[ ]

{ }

f ( ) y = exp m y −exp m y

  • ∞ < y < ∞

( ) { }

( ) { }

( ) exp exp
1 exp exp
F y y
y

(maximum)

(minimum)

σ

ˆ

β μ σ

μ σ

=

= +

(maximum)

(minimum)

(Double Exponential

Distribution)

The annual peak flood of a stream exceeds 2000m

3

/s

with a probability of 0.02 and exceeds 2250m

3

/s with a

probability of 0.

  1. Obtain the probability that annual peak flood exceeds

2500m

3

/s

Solution:

The parameters α and β are obtained from the given data

as follows

P[X > 2000] = 0.
P[X < 2000] = 0.

i.e., = 0.

Example-

(Gumbel s Extreme Value distribution)

4

( ) e

y

e

F y

  • ∞ < y < ∞

e

y

e

Solving (1) and (2),

α = 358 and β = 603

Now P[X > 2500] = 1 – P[X < 2500]

= 1 – exp{-exp(-y)}

y = (x – β)/ α

P[X > 2500] = 1 – exp{-exp(-5.299)}

Example-1 (contd.)

6

Extreme Value Type-III Distribution

  • Referred as Weibull distribution for minimum values
  • pdf is given by (for minimum values)
  • CDF is given by
  • Mean and variance of the distribution are

μ = E[X] = β Γ(1+1/α)

σ

2

= Var(X) = β

2

{Γ(1+2/α) – Γ

2

(1+1/α)}

7

( )

{ }

1

f ( ) x x exp x x 0; , 0

α

α α

α β β α β

− −

( ) { }

F x ( ) 1 exp x x 0; , 0

α

= − − β ≥ α β >

Weibull Distribution

  • Y = {(X – ε)/ (β – ε)}

α

→ transformation

  • Mean and variance of the 3-parameter Weibull

distribution are

μ = E[X] = ε + (β - ε) Γ(1+1/α)

σ

2

= Var(X) = (β- ε)

2

{Γ(1+2/α) – Γ

2

(1+1/α)}

9

Weibull Distribution

10

α → Shape parameter

β → Scale parameter

α=

β=

x

f(x)

α=

β=

α=

β=

α=

β=

α=

β=

Example-2 (contd.)

12

σ

2

= Var(X) = β

2

{Γ(1+2/α) – Γ

2

(1+1/α)}

2

2 2

2

2 2

2 2 2

×

β

α α

μ

α

μ = β Γ(1+1/α)

μ /β = Γ(1+1/α)

Substituting the

sample moment

values

P[X < 0.1] = F(0.1)
P[X < 0.1] = 0.

Example-2 (contd.)

13

( )

{ }

( )

{ }

( ) 1 exp
(0.1) 1 exp 0.1 0.
F x x
F

α

Parameter Estimation

  • f(x) : pdf and F(x) : CDF
  • In general, f(x) and F(x) are also functions of parameters

f(x; θ

1

; θ

2

………… θ

m

) or F(x; θ

1

; θ

2

………… θ

m

  • A random sample x

1

, x

2

, …… x

n

is available

: Estimate of θ

i

: a function of the sample

  • : random variable since it is a function of the random

sample.

  • Two important properties of the estimators

§ Unbiasedness

§ Consistency

15

i

i

Parameter Estimation

Unbiased estimate:

  • An estimate of a parameter θ is said to be unbiased if

E( ) = θ.

  • The bias is given by E( ) – θ
  • An estimator is unbiased does not guarantee that an

individual is equal to θ or even close to θ

  • The average of many independent estimates of θ will be

equal to θ.

16

Parameter Estimation

Methods of estimating parameters from samples of data:

  • Method of matching points:
  • Method of moments
  • Method of maximum likelihood

18

Method of matching points

  • Not a commonly used method
  • Can produce reasonable first approximations to the

parameters

  • Use the data set to obtain probabilities and estimate the

parameters

  • Simple and approximate method

19

Method of Moments (MoM)

  • One of the most common used methods for estimating

the parameters

  • Equate the first m moments of the population to the

sample estimates of the first m moments

  • Results in m equations; solve to get the m

unknown parameters of the distribution.

21

Example-

Obtain the parameter λ using method of moments for

the pdf

The first moment is

22

x

f x e x

λ

0

x

x e dx

λ

0 0

2

0

0

x x

x x

x

xe e

dx

xe e

e x

λ λ

λ λ

λ

∞ − −

− −