correlation & dependency structurescorrelation as the measure of dependence. • the mean does...

Correlation & DependencyStructures

GIRO - October 1999

Andrzej Czernuszewicz

Dimitris Papachristou

Why are we interested incorrelation/dependency?

• Risk management

• Portfolio management

• Reinsurance purchase

• Pricing

How does dependency arise in theinsurance world?

Related Events

• insurance cycle

• economic factors

• physical events

• social trends

• reinsurance failure

Items to be covered:

1. Simulations of correlated variables

2. Problems with the traditional approach

3. Copulas

Statement of problem

Suppose we have n classes of business witheach class of business having its ownmarginal distribution.

How do we model the portfolio?

Traditionally this problem is tackled usingcorrelation as the measure of dependence.

• The mean does not tell everything abouta distribution.

• The coefficient of correlation does nottell everything about the dependencystructure.

Standard Simulation Technique

Step 1

Simulate an n-dimensional multivariate normal distributionwith correlation matrix ρ

Simulate Multivariate Normal Variables

Simulated Independent Normal

-5

-4

-3

-2

-1

0

1

2

3

4

-4 -3 -2 -1 0 1 2 3 4

N1

N2

Simulated Multivariate Normal

-5

-4

-3

-2

-1

0

1

2

3

4

-4 -3 -2 -1 0 1 2 3 4

N 1

Step 2generate n series of the appropriate marginal distributions andput into a matrix

Step 3

Apply Normal multivariate dependency structure from step 1

Standard Simulation Technique

Simulated Independent Gamma Variables

0

0.5

1

1.5

2

2.5

3

3.5

4

0 0.5 1 1.5 2 2.5 3 3.5 4

X

Y

Simulated Multivariate Normal

-5

-4

-3

-2

-1

0

1

2

3

4

-4 -3 -2 -1 0 1 2 3 4

N 1

0

0.5

1

1.5

2

2.5

3

3.5

4

0 0.5 1 1.5 2 2.5 3 3.5 4X

Y

Simulated Independent Normal

-5

-4

-3

-2

-1

0

1

2

3

4

-4 -3 -2 -1 0 1 2 3 4

N1

N2

Standard Simulation Technique

Observations

• The Spearman rank correlations are matchedrather then the Pearson correlations

• The solution is not unique

• In particular use of normal distributioninfluences the tail of the modelled portfolio

The same Correlation, butDifferent Dependency Structures

0

0.5

1

1.5

2

2.5

3

3.5

4

0 0.5 1 1.5 2 2.5 3 3.5 4

X

Y

number of points in uper quartile = 9

0

0.5

1

1.5

2

2.5

3

3.5

4

0 0.5 1 1.5 2 2.5 3 3.5 4

X

Y

number of points in uper quartile = 23

Observation

In the majority of applications a symmetricapproach is used in determining the dependency.

However in practice this need not be the case.

Example: An earthquake may cause a stockmarketcrash but a stockmarket crash will not cause anearthquake.

Skewed distributions may be used to get aroundthis problem.

Asymmetric Dependency Structure

0

0.5

1

1.5

2

2.5

3

3.5

4

0 0.5 1 1.5 2 2.5 3 3.5 4

X

Ynumber of points in uper quartile =37

The same Correlation, butDifferent Dependency Structures

0

0.5

1

1.5

2

2.5

3

3.5

4

0 0.5 1 1.5 2 2.5 3 3.5 4X

Y

Two FallaciesFallacy 1

Marginal distributions and correlationsdetermine the joint distribution

Fallacy 2 (not for rank correlation)

F1, F2 marginal distribution for X1, X2

∀ ∈ [-1,1] ∃ F such that F is the jointdistribution and X1, X2 have Pearsoncorrelation

ρ

ρ

Other problems with Pearson correlation

• A correlation of zero does not indicateindependence of risk.

Problem 2 (not for rank)

• Correlation is not invariant undertransformations of the risk.

Problem 3

• Correlation is not an appropriate dependencemeasure for very heavily-tailed distributions.

Problem 1

Why dependency structures?

We need to amend our concept ofdependency to allow for desirable features

• In particular we have to introduce non-lineardependency. For example:

• Need to reflect special features of taildependence.

• In general, single numeric measures ofdependency are insufficient.

0,],1,1[u~ 2 ==− ρXYX

The Copula approach

• Operates by separating marginal distributionsfrom dependency structures

• Combination of copula and marginal will yieldoriginal distribution exactly

• No longer have the problems associated withcorrelation

Definition of Copula For m-variate distribution F with j th univariant margin Fj

the copula associated with F is a distribution function

(Note: if F is a continuous m-variate distribution the copula associated with F is unique)

))(),..,(()(

satisfiesthat]1,0[]1,0[:

11 mm

m

XFXFCXF

C

=→

This can also be represented via a density function

1,0,),(

),( <<∂∂

∂= vuvu

vuCvuc

Construction of a Copula

• Construction of a copula– parametric

– non-parametric

• parametric form needed for higher dimensionproblem

Example of Parametric Copula (1)

Independent Copula

1,0,1),(

1,0,),(

≤≤=≤≤=

vuvuc

vuuvvuC

Example of Parametric Copula (2)

Normal copula

ncorrelatiotheisand

)1,0(is)(),(

where

]}[2

1exp{]}.2[)1(

2

1exp{)1(),,(

11

2222122

12

ρ

ρρρρ

Nvyux

yxxyyxvuc

ΦΦ=Φ=

+−+−−−=

−−

−−

Example of Parametric Copula (3)

Gumbel Copula

For ∞<≤δ1

{ } 1,0,))log()log((exp);,(1

≤≤−+−−= vuvuvuC δδδδ

Note: this is an extreme value copula

Independent (Product) Copula Density

0.0

1

0.1

0.1

9

0.2

8

0.3

7

0.4

6

0.5

5

0.6

4

0.7

3

0.8

2

0.9

1 1

0.01

0.2

0.39

0.58

0.77

0.96

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Density

X

Y

Independent (Product) Copula

0.9-1

0.8-0.9

0.7-0.8

0.6-0.7

0.5-0.6

0.4-0.5

0.3-0.4

0.2-0.3

0.1-0.2

0-0.1

0.0

05

0.0

75

0.1

45

0.2

15

0.2

85

0.3

55

0.4

25

0.4

95

0.5

65

0.6

35

0.7

05

0.7

75

0.8

45

0.9

15

0.9

85

0.005

0.185

0.365

0.545

0.725

0.905

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2

X

Y

Bivariate Normal Copula (Density)

0.18-0.2

0.16-0.18

0.14-0.16

0.12-0.14

0.1-0.12

0.08-0.1

0.06-0.08

0.04-0.06

0.02-0.04

0-0.02

Bivariate Normal Copula (Density)

0.0

1

0.0

8

0.1

5

0.2

2

0.2

9

0.3

6

0.4

3

0.5

0.5

7

0.6

4

0.7

1

0.7

8

0.8

5

0.9

2

0.9

9

0.01

0.19

0.37

0.55

0.73

0.91

-1

1

3

5

7

9

11

13

15

17

19

21

23

25

27

29

31

33

35

Copula Density Function

X

Y

Gumbel Copula (Density)

33-35

31-33

29-31

27-29

25-27

23-25

21-23

19-21

17-19

15-17

13-15

11-13

9-11

7-9

5-7

3-5

1-3

-1-1

Gumbel Copula (Density)

Simulation of a copula

• simulate a value u1 from U (0,1)

• simulate a value u2 from C2 (u2 u1)

• simulate a value un from Cn (un u1…. un-1)

where Ci = C(u1,….,ui,1,….,1) for i=2,….,n

• Stop Loss on two classes of business

• Assumptions

• Limit: 15%

• Deductible: 107.5%

• Expected L/R: 91.5%

Example: Reinsurance Pricing

Class A Class Bmean L/R 91.14% 91.85%st. dev. of L/R 10.98% 9.21%Premium 100 100

gumbel delta 1.2rank correlation 0.25

Distribution of Loss Ratios for Classes A and B

0%

10%

20%

30%

40%

50%

60%

70%

80%

90%

100%

50% 70% 90% 110% 130% 150% 170%

Loss Ratio

Cu

mu

lati

ve D

istr

ibu

tio

n

Class A

Class B

Loss Ratios are assumed to be lognormally distributed

The same rank correlation ( =0.25), but differentdependency structures especially at the tail

ρ

Distribution of Losses to the Stop Loss Layer

95%

96%

97%

98%

99%

100%

0 5 10 15 20 25

Amount of Loss

Cu

mu

lati

ve D

istr

ibu

tio

n

gumbel dependence

independent

MN dependence

Distribution of Losses to the Stop Loss Layer

Results

Expected amount of loss to the layer isunderestimated approximately by 50%, if amultivariate Normal dependency structure isused.Losses to the Stop Loss

Gumbel Independent Mult.Normalmean 0.325 0.116 0.220standard deviation 2.297 1.169 1.735Rate on Line 1.1% 0.4% 0.7%

Conclusions

• Correlation is not a sufficient measure ofdependence.

• Opportunities may be missed by remaining inthe correlation framework.

• True dependency reflected in the copulaapproach by separating marginal distributionfrom dependency structures.