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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
Karolina Wojciechowska (TU Delft) Dependency models with bivariate case study July 26, 2007 2 / 30
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.
Karolina Wojciechowska (TU Delft) Dependency models with bivariate case study July 26, 2007 3 / 30
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 ≤ ∞
Karolina Wojciechowska (TU Delft) Dependency models with bivariate case study July 26, 2007 4 / 30
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
Karolina Wojciechowska (TU Delft) Dependency models with bivariate case study July 26, 2007 5 / 30
Estimation - Maximum Likelihood Inference∏n
i=1{c(F (Xi ), G(Yi ))f (Xi )g(Yi )}
Karolina Wojciechowska (TU Delft) Dependency models with bivariate case study July 26, 2007 6 / 30
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)
Karolina Wojciechowska (TU Delft) Dependency models with bivariate case study July 26, 2007 7 / 30
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.
Karolina Wojciechowska (TU Delft) Dependency models with bivariate case study July 26, 2007 8 / 30
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]}
Karolina Wojciechowska (TU Delft) Dependency models with bivariate case study July 26, 2007 9 / 30
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.
Karolina Wojciechowska (TU Delft) Dependency models with bivariate case study July 26, 2007 10 / 30
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.
Karolina Wojciechowska (TU Delft) Dependency models with bivariate case study July 26, 2007 11 / 30
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]
Karolina Wojciechowska (TU Delft) Dependency models with bivariate case study July 26, 2007 12 / 30
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
Karolina Wojciechowska (TU Delft) Dependency models with bivariate case study July 26, 2007 13 / 30
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)
Karolina Wojciechowska (TU Delft) Dependency models with bivariate case study July 26, 2007 14 / 30
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))
Karolina Wojciechowska (TU Delft) Dependency models with bivariate case study July 26, 2007 15 / 30
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))
Karolina Wojciechowska (TU Delft) Dependency models with bivariate case study July 26, 2007 16 / 30
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)
Karolina Wojciechowska (TU Delft) Dependency models with bivariate case study July 26, 2007 17 / 30
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.
Karolina Wojciechowska (TU Delft) Dependency models with bivariate case study July 26, 2007 18 / 30
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)
Karolina Wojciechowska (TU Delft) Dependency models with bivariate case study July 26, 2007 19 / 30
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.
Karolina Wojciechowska (TU Delft) Dependency models with bivariate case study July 26, 2007 20 / 30
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)
Karolina Wojciechowska (TU Delft) Dependency models with bivariate case study July 26, 2007 21 / 30
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).
Karolina Wojciechowska (TU Delft) Dependency models with bivariate case study July 26, 2007 22 / 30
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
Karolina Wojciechowska (TU Delft) Dependency models with bivariate case study July 26, 2007 23 / 30
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.
Karolina Wojciechowska (TU Delft) Dependency models with bivariate case study July 26, 2007 24 / 30
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.
Karolina Wojciechowska (TU Delft) Dependency models with bivariate case study July 26, 2007 25 / 30
Rotation Model - construction
Model (S , T )
fS,T (s, t) = e−sλσ(t), s ≥ 0
λσ - density function
σ > 0 - standard deviation
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
Karolina Wojciechowska (TU Delft) Dependency models with bivariate case study July 26, 2007 26 / 30
Rotation Model - construction
Rotate space (X , Y ) to (S , T )
fX ,Y (x , y) = e− x+y
√
2 λσ
(
−x+y√2
)
λσ - density function
σ > 0 - standard deviation
−3
−2
−1
0
1
2
3−1
0
1
2
3
4
5Rotation model
x
y
s
t
Karolina Wojciechowska (TU Delft) Dependency models with bivariate case study July 26, 2007 26 / 30
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).
Karolina Wojciechowska (TU Delft) Dependency models with bivariate case study July 26, 2007 27 / 30
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
Karolina Wojciechowska (TU Delft) Dependency models with bivariate case study July 26, 2007 28 / 30
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?).
Karolina Wojciechowska (TU Delft) Dependency models with bivariate case study July 26, 2007 29 / 30
Thank you!
Karolina Wojciechowska (TU Delft) Dependency models with bivariate case study July 26, 2007 30 / 30
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