Master’s thesis presentation Karolina Wojciechowska

Dependency models with bivariate case studyMaster’s thesis presentation

Karolina Wojciechowska

TU Delft

July 26, 2007

Karolina Wojciechowska (TU Delft) Dependency models with bivariate case study July 26, 2007 1 / 30

Overview

1 Copula functionBasic factsArchimedean class of copulasStatistical inferenceCase study

2 Bivariate conditional modelsTransformationConstant Spread ModelVariable Spread ModelConstant Symmetric Spread ModelCase study

3 Rotation ModelConstructionPropertiesCase study

4 Conclusions and recommendations


Copula function - basic facts

A copula is a distribution function defined on the unit square with uniformmarginal distributions.

Sklar’s theorem H(x , y) = C (F (x), G(y)).

If F and G are continuous then C is unique.

h(x , y) = c(F (x), G(y))f (x)g(y)

A copula describes margin-free dependence between random variables.


Archimedean class of copulas - definition

DefinitionA bivariate copula C is called Archimedean with generator φ if:

C (u, v) = φ[−1](φ(u) + φ(v))

a generator φ : [0, 1] → [0,∞] is convex, strictly decreasing φ(1) = 0 and itspseudo-inverse φ[−1] is:

φ[−1](t) =

{

φ−1(t) 0 ≤ t ≤ φ(0)0 φ(0) ≤ t ≤ ∞


Archimedean class of copulas - examples

00.5

1

0

0.5

10

2

4

6

8

10

12

Clayton denisty θ=2

0

0.5

1

0

0.5

10

1

2

3

4

5

6

7

8

Gumbel denisty θ=2

00.5

1

0

0.5

10.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

Frank denisty θ=2


Estimation - Maximum Likelihood Inference∏n

i=1{c(F (Xi ), G(Yi ))f (Xi )g(Yi )}


Pseudo-graphical method for Archimedean class

Assume that copula C is Archimedean and consider a sample of n bivariateobservations {(Xi , Yi )}

i=ni=1. The method is based on comparison of

parametric and non-parametric estimations of functionK (t) = P{C (U, V ) ≤ t}, for t ∈ [0, 1].

The non-parametric estimation of K (t)

Wi =

n∑

j=1

1(Xj ≤ Xi , Yj ≤ Yi )/n ⇒ Kn(t) =

n∑

i=1

1(Wi ≤ t)/n

The parametric estimation of K (t)

θ̂ ⇒ K (t; θ̂) = P{C (U, V ; θ̂) ≤ t} = t −φ(t, θ̂)

φ′(t+, θ̂)


Pseudo-graphical method for Archimedean class

Plot Kn(t) and K (t; θ̂) on one graph - if the plots ”agree” then the copulafits well.

Compute distance E :

E =

∫ 1

0

|Kn(t) − K (t; θ̂)|2dt

The best copula minimizes this distance.

Perform a statistical test with the test statistic:

Sn = n

∫ 1

0

|Kn(t) − K (t; θ̂)|2k(t; θ̂)dt

The Bootstrap method is used to obtain the critical value.


Method of percentile lines

(X , Y ) - physical space

(F (X ), G(Y )) - copula space

DefinitionThe p-percentile line in the copula space is:

v = f (u; p), u ∈ [0, 1]

where p is a percentage and f (u; p) solves theequation C (v |u) = p.

p-percentile line in the physical space:

{(u, f (u; p)) : u ∈ [0, 1]} ⇒ {(F−1(u), G−1(f (u; p))) : u ∈ [0, 1]}


Case study - dataset

Covariance matrix:

Σ =

[

0.26 0.610.61 7.50

]

0 0.5 1 1.5 2 2.5 36

8

10

12

14

16

18

20

Water level

Win

d sp

eed

Dataset

Figure: 89 observations of water level and wind speed.


Case study - marginal distributions

0 1 2 310

−3

10−2

10−1

100

The exceedence prob. of the water level

V

1−F

V(v

)

5 10 15 2010

−3

10−2

10−1

100

The exceedence prob. of the wind speed

W

1−F

W(w

)

New distr. (data)Empirical distr. (data)Weibull distr.

Figure: The functions 1 − FV (v) and 1 − FW (w) on a logarithmic scale.


Fitting copula

Consider three bivariate Archimedean copulas: Clayton, Gumbel and Frank.

Estimated parameters and 95% confidence intervals:

Copula θ̂ 95% confidence interval

Clayton 0.42 [0.10, 0.74]Gumbel 1.44 [1.21, 1.67]Frank 2.82 [1.50, 4.14]


Evaluation of the fit - comparison of Kn(t) and K (t; θ̂)

0 0.5 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

t

Kn(t

) an

d K

(t)

Clayton copula

0 0.5 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

t

Gumbel copula

0 0.5 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

t

Frank copula

Clayton copula

E p-value

0.00195 0.106

Gumbel copula

E p-value

0.00094 0.65

Frank copula

E p-value

0.0017 0.214


Evaluation of the fit - percentile lines

0 1 2 3 46

8

10

12

14

16

18

20Original space

V

W

0 0.2 0.4 0.6 0.8 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1Copula space−Clayton copula

FV(V)

FW

(W)

0 1 2 3 46

8

10

12

14

16

18

20

22Original space

V

W

0 0.2 0.4 0.6 0.8 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1Copula space−Gumbel copula

FV(V)

FW

(W)

0 1 2 3 46

8

10

12

14

16

18

20Original space

V

W

0 0.2 0.4 0.6 0.8 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1Copula space−Frank copula

FV(V)

FW

(W)


Transformation

Consider random vector (V , W ) with known FV and FW , and unknown fV ,W .

Consider model (X , Y ) with known FX , FY and fX ,Y .

Transform (V , W ) to (X , Y ):

Transformation

x(v) = F−1X (FV (v)) and y(w) = F−1

Y (FW (w))

Then model fV ,W is:

fV ,W (v , w) =fX ,Y (x(v), y(w))fV (v)fW (w)

fX (x(v))fY (y(w))


Transformation - example

Consider model (X , Y ) with FX (x) = 1 − e−x and FY (y) = 1 − e−y andsome fX ,Y .

0 1 2 36

8

10

12

14

16

18

20

V

W

Original space

0 1 2 3 4 50

0.5

1

1.5

2

2.5

3

3.5

4

4.5

XY

Transformed space

Transformation

x=F−1X

(FV(v))

y=F−1Y

(FW

(w))


Constant Spread Model (Model CS) - construction

Model (X , Y )

fX (x) = e−x , x ≥ 0

fY |X=x(y |x) = λσ(y − x − δ)

λσ - density function

E (Y |X = x) = x + δ

σ > 0 - standard deviation

fX ,Y (x , y) = e−xλσ(y − x − δ)

fY (y) - ”exponential” −1 0 1 2 3 4 5−3

−2

−1

0

1

2

3

4

5

6

7

y

x

Model CS−infinite support

fY|X=x

(y|x)=λσ(y−x−δ)y=x+δ

fX(x)=e−x

fY(y)


Model CS - tail dependence and related copula

Lower tail dependence coefficient

λL = limu↓0

P{Y ≤ F−1Y (u)|X ≤ F−1

X (u)}

Upper tail dependence coefficient

λU = limu↑1

P{Y > F−1Y (u)|X > F−1

X (u)}

λL = 0

If λσ is a normal density then λU = 2Φ(−σ/2), Φ - cdf of standard normalvariable.

If λσ is a normal density then the related copula does not belong to theArchimedean class.


Variable Spread Model (Model VS) - construction

Model (X , Y )

fX (x) = e−x , x ≥ 0

fY |X=x(y |x) = λσs (x)(y − x)

λσs (x) - density function

E (Y |X = x) = x

σs(x) > 0 - ”spread function”

fX ,Y (x , y) = e−xλσs (x)(y − x)−1 0 1 2 3 4 5

−2

−1

0

1

2

3

4

5

6

7

y

x

Model VS

fY|X=x

(y|x)=λσ(x)(y−x) y=x

fX(x)=e−x

fY(y)


Model VS - tail dependence and related copula

If λσs (x) is a normal density and σs(x) > 0 is increasing andlimx→∞ σs(x) = σ1 < ∞:

λL = 0 and λU = 2Φ(−σ1/2)

If λσs (x) is a normal density and σs(x) = 2 + 0.5x then the related copuladoes not belong to the Archimedean class.


Constant Symmetric Spread Model - construction

Model (X , Y )

fX (x), fY (y) - assym. exp.

fY |X=x(y |x) - modified normal

σ > 0 - spread

−1 0 1 2 3 4−2

−1

0

1

2

3

4

5

6

7

x

y

Model CSS

fY|X=x

(y|x)

y=x−σ2/2

fX(x)

fY(y)


Model CSS - tail dependence and related copula

The lower tail independence occurs, because λL = 0.

The upper tail dependence occurs, because λU = 2Φ(−σ/2).

The related copula function is ”close” to the Archimedean class (numericalexperiments).


Model CS - case study

Assume that λσ is a normal density, δ = 0, the unknown parameter is σ.

0 5 10−4

−2

0

2

4

6

8

10

12

14

16Transformed space−Model CS

X

Y

0 2 45

10

15

20

25

30

35Original space

V

W

0 0.5 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Fv

Fw

Copula space

Figure: Model CS, σ̂ = 1.6


Model VS - case studyAssume that λσs (x) is a normal density and the spread function σs(x) is:

σs(x) =

{

σ

5 x + σ for x ∈ [0, 5]2σ for x > 5

0 5 10−4

−2

0

2

4

6

8

10

12

14Transformed space−Model VS

X

Y

0 1 2 3 45

10

15

20

25

30Original space

V

W

0 0.5 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

FvF

w

Copula space

Figure: Model VS, σ̂ = 1.1.


Model CSS - evaluation of the fit

0 5 10−2

0

2

4

6

8

10

12

14Transformed space−Model CSS

X

Y

0 2 45

10

15

20

25

30

35Original space

V

W

0 0.5 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Fv

Fw

Copula space

Figure: Model CSS, σ̂ = 1.4.


Rotation Model - construction

Model (S , T )

fS,T (s, t) = e−sλσ(t), s ≥ 0



0 1 2 3 4 5−4

−3

−2

−1

0

1

2

3

4

s

t

Basic model

fT|S=s

(t|s)=λσ(t)

fS(s)=e−s


Rotation Model - construction

Rotate space (X , Y ) to (S , T )

fX ,Y (x , y) = e− x+y

√

2 λσ

(

−x+y√2

)



−3

−2

−1

0

1

2

3−1

0

1

2

3

4

5Rotation model

x

y

s

t


Rotation Model - properties

If λσ has finite support then fX (x) and fY (y) become shifted exponential.

If λσ is a normal density then fY |X=x(y |x) is a modified normal density.

The definition of the model always entails lower tail independence λL = 0.

If λσ is a normal density then λU = 2Φ(−σ).


Case study - problem

0 1 2 36

8

10

12

14

16

18

20The original space

V

W

−5 0 5−3

−2

−1

0

1

2

3

4

5The transformed space

X

Y

Transformation

fV,W

(v,w)=0 fX,Y

(x,y)=0


Conclusions and recommendations

Gumbel copula is a good model to the considered dataset.

It is a bit difficult to judge the fit in the extreme region in the copula spaceusing the percentile lines method.

The lower tail independence and upper tail dependence occur for theconsidered conditional models.

It is relatively easy to judge the fit in the extreme region in the model spaceusing the percentile lines method.

The Rotation Model is not always suitable.

Further work with evaluation of the fit (AIC coefficient for models?).

Further work with the Rotation Model (Some additional conditions?).


Thank you!


Master’s thesis presentation Karolina Wojciechowska

Documents

Transcript of Master’s thesis presentation Karolina Wojciechowska