# Copula Regression

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05-Feb-2016Category

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### Transcript of Copula Regression

BYRAHUL A. PARSADRAKE UNIVERSITY

&STUART A. KLUGMANSOCIETY OF ACTUARIES

Copula Regression

Outline of TalkOLS RegressionGeneralized Linear Models (GLM)Copula RegressionContinuous caseDiscrete CaseExamples

NotationNotation:Y Dependent Variable AssumptionY is related to Xs in some functional form

OLS RegressionY is linearly related to XsOLS Model

OLS RegressionEstimated Model

OLS Multivariate Normal DistributionAssume Jointly follow a multivariate normal distributionThen the conditional distribution of Y | X follows normal distribution with mean and variance given by

OLS & MVNY-hat = Estimated Conditional mean

It is the MLE

Estimated Conditional Variance is the error variance

OLS and MLE result in same values

Closed form solution exists

GLMY belongs to an exponential family of distributions

g is called the link functionx's are not random Y|x belongs to the exponential familyConditional variance is no longer constantParameters are estimated by MLE using numerical methods

GLMGeneralization of GLM: Y can be any distribution (See Loss Models)

Computing predicted values is difficult

No convenient expression conditional variance

Copula RegressionY can have any distribution

Each Xi can have any distribution

The joint distribution is described by a Copula

Estimate Y by E(Y|X=x) conditional mean

CopulaIdeal Copulas will have the following properties:ease of simulation closed form for conditional density different degrees of association available for different pairs of variables.

Good Candidates are:Gaussian or MVN Copulat-Copula

MVN CopulaCDF for MVN is Copula is

Where G is the multivariate normal cdf with zero mean, unit variance, and correlation matrix R.Density of MVN Copula is

Where v is a vector with ith element

Conditional Distribution in MVN CopulaThe conditional distribution of xn given x1 .xn-1 is

Where

Copula RegressionContinuous Case

Parameters are estimated by MLE.

If are continuous variables, then we use previous equation to find the conditional mean.one-dimensional numerical integration is needed to compute the mean.

Copula RegressionDiscrete CaseWhen one of the covariates is discreteProblem:determining discrete probabilities from the Gaussian copula requires computing many multivariate normal distribution function values and thus computing the likelihood function is difficultSolution:Replace discrete distribution by a continuous distribution using a uniform kernel.

Copula Regression Standard ErrorsHow to compute standard errors of the estimates?As n -> , MLE , converges to a normal distribution with mean q and variance I(q)-1, where

I(q) Information Matrix.

How to compute Standard ErrorsLoss Models: To obtain information matrix, it is necessary to take both derivatives and expected values, which is not always easy. A way to avoid this problem is to simply not take the expected value.

It is called Observed Information.

ExamplesAll examples have three variables

R Matrix :

Error measured by

Also compared to OLS

10.70.70.710.70.70.71

Example 1Dependent X3 - GammaThough X2 is simulated from Pareto, parameter estimates do not converge, gamma model fit

Error:

VariablesX1-ParetoX2-ParetoX3-GammaParameters3, 1004, 3003, 100MLE3.44, 161.111.04, 112.0033.77, 85.93

Copula59000.5OLS637172.8

Ex 1 - Standard ErrorsDiagonal terms are standard deviations and off-diagonal terms are correlations

Sheet1

R(2,1)R(3,1)R(3,2)

0.2666060.9660670.359065-0.337250.349482-0.33268-0.42141-0.33863-0.29216

0.96606715.509740.390428-0.252360.346448-0.26734-0.37496-0.29323-0.25393

0.3590650.3904280.025217-0.787660.438662-0.35533-0.45221-0.30294-0.42493

-0.33725-0.25236-0.787663.558369-0.384890.4645130.4968530.356080.470009

0.3494820.3464480.438662-0.384890.100156-0.93602-0.34454-0.46358-0.46292

-0.33268-0.26734-0.355330.464513-0.936022.4853050.3656290.4821870.481122

R(2,1)-0.42141-0.37496-0.452210.496853-0.344540.3656290.0100850.4574520.465885

R(3,1)-0.33863-0.29323-0.302940.35608-0.463580.4821870.4574520.010080.481447

R(3,2)-0.29216-0.25393-0.424930.470009-0.462920.4811220.4658850.4814470.009706

Sheet2

Sheet3

Example 1 - ContMaximum likelihood Estimate of Correlation MatrixR-hat =

10.7110.6990.71110.7130.6990.7131

Example 2Dependent X3 - GammaX1 & X2 estimated Empirically

Error:

VariablesX1-ParetoX2-ParetoX3-GammaParameters3, 1004, 3003, 100MLEF(x) = x/n 1/2n f(x) = 1/nF(x) = x/n 1/2n f(x) = 1/n4.03, 81.04

Copula595,947.5OLS637,172.8GLM814,264.754

Example 3Dependent X3 - GammaPareto for X2 estimated by Exponential

Error:

VariablesX1-PoissonX2-ParetoX3-GammaParameters54, 3003, 100MLE5.65119.393.67, 88.98

Copula574,968OLS582,459.5

Example 4Dependent X3 - GammaX1 & X2 estimated EmpiricallyC = # of obs x and a = (# of obs = x)

Error:

VariablesX1-PoissonX2-ParetoX3-GammaParameters54, 3003, 100MLEF(x) = c/n + a/2nf(x) = a/nF(x) = x/n 1/2n f(x) = 1/n3.96, 82.48

CopulaOLSGLM559,888.8582,459.5652,708.98

Example 5Dependent X1 - PoissonX2, estimated by Exponential

Error:

VariablesX1-PoissonX2-ParetoX3-GammaParameters54, 3003, 100MLE5.65119.393.66, 88.98

Copula108.97OLS114.66

Example 6Dependent X1 - PoissonX2 & X3 estimated by Empirically

Error:

VariablesX1-PoissonX2-ParetoX3-GammaParameters54, 3003, 100MLE5.67F(x) = x/n 1/2n f(x) = 1/nF(x) = x/n 1/2n f(x) = 1/n

Copula110.04OLS114.66

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