Archimedean Copulas Theodore Charitos MSc. Student CROSS.

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Archimedean Copulas Theodore Charitos MSc. Student CROSS

Transcript of Archimedean Copulas Theodore Charitos MSc. Student CROSS.

Page 1: Archimedean Copulas Theodore Charitos MSc. Student CROSS.

Archimedean Copulas

Theodore Charitos

MSc. Student

CROSS

Page 2: Archimedean Copulas Theodore Charitos MSc. Student CROSS.

Task-Goal

• Examination of relations between Archimedean copulas and diagonal band or minimum information copulas given correlation constraints

• Computation of relative information with respect to uniform distribution for each family.

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Accomplishment of tasks

• Use of the algorithm provided in the paper of Christian Genest and Louis-Paul Rivest “Statistical Inference Procedures for Bivariate Archimedean Copulas”.

• Use of small program in Matlab for calculating numerically the relative information with respect to uniform distribution.

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Structure of presentation

• Theoretical Background

• Explanation and description of the whole procedure proposed by Genest and Rivest.

• Analysis of datasets sampled from Unicorn software.

• Results

• Conclusions-Discussions

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Theoretical Background

Definition: A bivariate distribution function with marginals and is said to be generated by an Archimedean copula if it can be expressed in the form

for some convex,

decreasing function on in such a way that

xF yG

yxH , yGxF 1

1,0 01

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Proposition: Let and be uniform random variates whose

dependence function is of the form

for some convex decreasing function defined on with the

property that . Set ,

and . Then is uniformly distributed on ,

is distributed as and , are

independent random variables.

X Y

yxH , yx 1

1,0

01

YX

XU

YXHV ,

u

uu

' U 1,0

uuuK UV V

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Proposition: Let X and Y be uniform random variables with dependence function . For let

and define

The function is convex,decreasing and satisfies if

and only if for all .

It is obvious from the above propositions that is determined as long as can be determined from the dataset. This will be done in our case via a nonparametric estimation of the distribution

of V based on a decomposition of Kendall’s tau.

yxH , 10 u

uYXHuK ,Pr tKuKut

lim

u 01

uuK 10 u

u u

uK

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A pair of random variables is concordant if large values of one tend to be associated with large values of the other and vice versa. More precisely, if we have two observations

and from a vector of continuous random variables we say that and

are concordant if and . Similarly,

and are discordant if

and or vice versa.

ii yx ,

jj yx , YX , ii yx , jj yx ,

ji xx ji yy

ii yx , jj yx , ji xx

ji yy

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DefinitionThe Kendall’s tau for the sample is defined as

where c is the number of concordant pairs, d is the number of discordant pairs from n observations of a vector and is the number of distinct pairs of observations in the sample.

2

n

dc

YX ,

2

n

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For Archimedean copulas the Kendall’s tau statistic can be conveniently computed via the identity

Apparently, the problem now of estimating the bivariate dependence function relies on the estimation of . Genest and Rivest provide a nonparametric procedure for estimating

and also .

1

0

1414 duuVE

uK

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Analysis of various datasetsThe algorithm proposed by Genest and Rivest uses

the variables where

the symbol # stands for the cardinality of a set. If

denotes the distribution function of a point mass at the origin, then a nonparametric estimator of is given by

Knowing that , a sample equivalent for the estimation of is

1

,:,#

n

YYXXYXV ijijjj

i

t

uK

ni

n n

VuuK

1

14 VE 14 Vn

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Family

Clayton

Frank

Gumbel

-

a

v a 1 a

vv a12a

a

av

a

exp1

exp1log

av

a

ava

av

exp1

exp1log

exp

exp1 a

aD 141 1

1log av

1

log

a

vv

1a

a

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The next step of the analysis concerns the performance of a Pearson chi-squared goodness of fit test statistic for each family in order to assess the fit of the various models. This means that a classification of the dataset is made each time constituting the observed frequencies. However, since the chi-squared test requires predicted values for its computation, it is necessary to generate random variates

whose joint distribution belongs to one of the mentioned Archimedean families

vu,

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Algorithm for sampling from Archimedean families

1.Generate two independent uniform variates

u and t.

2. Set

3. Set v =

4. The desired pair is

1,0

t

uw

'' 1

uw 1

vu,

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Archimedean Family

Clayton

Frank

Gumbel

yxH ,

aaa yx/1

1

1

111ln

1a

ayax

e

ee

a

1/111 lnlnexp

aaa yx

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Clayton’s joint density with a=1.514 Frank’s joint density with a=4.604

Gumbel’s joint density with a=0.757

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In general, the relative information with respect to uniform distribution for the bivariate case is

computed as

where is the joint density of and

An approximation of the real solution in each case will be provided, which however is enough to indicate what should someone expect from each Archimedean family.

1

0

1

0

,log,/ dxdyyxhyxhuhI

yxh , X Y

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• To illustrate the above procedure six datasets (n=1000) were at first sampled and thoroughly analyzed. The correlations were 0.2, 0.65 and 0.9 for both the diagonal band and the minimum information copulas. A classification of the frequencies was also decided.

• For the sake of completeness, six more datasets with similar correlations constraints but different size (n=5000) and classification were also analyzed in order to compare results.

44

66

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Recapitulation-Steps

• Sample from diagonal band and minimum information copulas.

• Estimate Kendal’s tau and the empirical lambda function.

• Estimate the parameters for each family according to the previous results.

• Estimate the lambda functions for each family.

• Classify the dataset in categories and simulate values from each family according to their estimated parameters.

• Perform chi-square goodness of fit test and compare the resulting fits.

• Compute the relative information with respect to uniform distribution.

• Repeat the whole procedure for different correlations and size of the dataset.

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Examples of classifications from diag.band with 0.2 Cross-Classification of X and Y (Observed values) X\Y

8 43 36 0 47 166 144 38 40 131 186 52 0 49 46 14 Cross-Classification of X and Y (Observed values) X\Y

67 126 118 111 78 1 105 222 230 173 118 78 129 211 207 131 203 131 126 196 121 199 244 128 80 119 183 233 239 127 0 84 123 152 126 77

1.0,0

5.0,1.0 9.0,5.0

1,9.0

44 1.0,0 5.0,1.0 9.0,5.0 1,9.0

66 1.0,0 3.0,1.0 5.0,3.0 7.0,5.0 9.0,7.0 1,9.0

1.0,0

3.0,1.0

5.0,3.0

7.0,5.0

9.0,7.0

1,9.0

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=0.0877273

Statistic df

Clayton 47.2727 8

Frank 34.4113 8

Gumbel 49.5668 8

=0.1076116

Statistic df

Clayton 52.3819 7

Frank 32.7662 8

Gumbel 47.2727 8

n2

n2

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=0.431007

Statistic df

Clayton 219.842 5

Frank 66.528 7

Gumbel 129.861 6

=0.424140

Statistic df

Clayton 113.601 5

Frank 24.808 7

Gumbel 119.047 7

n

n2

2

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=0.6820981

Statistic df

Clayton 265.805 3

Frank 67.031 3

Gumbel 182.181 3

=0.6991151

Statistic df

Clayton 215.628 3

Frank 39.309 3

Gumbel 140.177 3

n

n2

2

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=0.1083165

Statistic df

Clayton 356.184 25

Frank 301.967 25

Gumbel 357.828 25

=0.1040581

Statistic df

Clayton 148.598 25

Frank 65.967 25

Gumbel 94.2128 25

n

n2

2

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=0.4270299

Statistic df

Clayton 1707.548 22

Frank 604.8511 23

Gumbel 1398.062 23

=0.4398811

Statistic df

Clayton 1044.165 23

Frank 127.982 23

Gumbel 557.023 23

n

n

2

2

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=0.6801317

Statistic df

Clayton 1685.918 15

Frank 473.3928 17

Gumbel 1128.659 17

=0.6906168

Statistic df

Clayton 1142.479 14

Frank 124.384 16

Gumbel 677.908 17

n

n2

2

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Relative Information with respect to uniform distribution

Clayton Frank Gumbel

Diag.band 0.2 (n=1000) 0.0144 0.8261 0.0887

Min.inf.0.2 (n=1000) 0.0215 0.7925 0.1099

Diag.band 065 (n=1000) 0.3207 0.4932 0.5222

Min.inf.0.65 (n=1000) 0.3107 0.4944 0.5126

Diag.band 0.9 (n=1000) 0.8653 0.6884 0.8794

Min.inf.0.9 (n=1000) 0.9202 0.7247 0.9015

Diag.band0.2 (n=5000) 0.0218 0.7913 0.1106

Min.inf.0.2 (n=5000) 0.0202 0.7983 0.1060

Diag.band 0.65 (n=5000) 0.3150 0.4939 0.5167

Min.inf.0.65 (n=5000) 0.3343 0.4920 0.5351

Diag.band0.9 (n=5000) 0.8592 0.6844 0.8768

Min.inf.0.9 (n=5000) 0.8924 0.7060 0.8905

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Conclusions-Comments

• for correlation 0.2 all the three families seem to fit reasonably well when n=1000,but when n=5000 only Frank’s and also Gumbel’s for the min.information.

• for correlation 0.65 the results are quite promising only when n=1000

• for correlation 0.9 Frank’s and Gumbel’s family seem to fit the data when n=1000 and only Frank’s family when n=5000.

• the results are more promising in the cases of minimum information copula and this is actually a fact that holds for all the datasets no matter what the correlation is.

• the results are much better when the size of the dataset is smaller.

• It is obvious that the chi-square test statistic is sensitive to the number of cells.For greater size n and 6x6 cross-classification the results in almost all cases are disappointing. A performance of another goodness of fit test might result in more encouraging conclusions.

• for correlation 0.2 Clayton’s family has the smallest values of relative information with respect to uniform distribution.

• Nonetheless, for greater correlations, Frank’s family has the smallest values.