master’s thesis presentation karolina wojciechowska
TRANSCRIPT
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Dependency models with bivariate case studyMaster’s thesis presentation
Karolina Wojciechowska
TU Delft
July 26, 2007
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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
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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.
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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 ≤ ∞
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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
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Estimation - Maximum Likelihood Inference∏n
i=1{c(F (Xi ), G(Yi ))f (Xi )g(Yi )}
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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)
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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.
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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]}
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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.
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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.
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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]
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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
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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)
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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))
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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))
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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)
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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.
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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)
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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.
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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)
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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).
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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
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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.
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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.
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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
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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
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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).
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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
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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?).
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Thank you!
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