On Predictive Modeling for Claim Severity Glenn Meyers ISO Innovative Analytics

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On Predictive Modeling for

Claim Severity

Glenn Meyers

ISO Innovative Analytics

CARe Seminar

June 6-7, 2005

Problems with

Experience Rating for

Excess of Loss Reinsurance

• Use submission claim severity data

– Relevant, but

– Not credible

– Not developed

• Use industry distributions

– Credible, but

– Not relevant (???)

General Problems with

Fitting Claim Severity Distributions

• Parameter uncertainty

– Fitted parameters of chosen model are estimates subject to sampling error.

• Model uncertainty

– We might choose the wrong model. There is no particular reason that the models we choose are appropriate.

• Loss development

– Complete claim settlement data is not always available.

Outline of Remainder of Talk

• Quantifying Parameter Uncertainty

– Likelihood ratio test

• Incorporating Model Uncertainty

– Use Bayesian estimation with likelihood functions

– Uncertainty in excess layer loss estimates

• Bayesian estimation with prior models based on data reported to a statistical agent

– Reflect insurer heterogeneity

– Develops losses

How Paper is Organized

• Start with classical hypothesis testing.

– Likelihood ratio test

• Calculate a confidence region for parameters.

• Calculate a confidence interval for a function of the parameters.

– For example, the expected loss in a layer

• Introduce a prior distribution of parameters.

• Calculate predictive mean for a function of parameters.

The Likelihood Ratio Test

Let p

 p

1 p k

) be a parameter vector for your chosen loss model.

Let x

 x

1 x n

) be a set of observed losses.

The Likelihood Ratio Test

Test H :

0 p

 p

*

against H :

1 p

 p

*

Theorem 2.10 in Klugman, Panjer & Willmot

If H is true then:

0

ln LR

2 ln

 

 ln

  has a

2 k of freedom.

Use

2 distribution to find critical values.

An Example – The Pareto Distribution

( ) 1

 x

  

• Simulate random sample of size 1000

= 2.000,

= 10,000

Maximum Likelihood = -10034.660 with

 ˆ 

8723.04

 ˆ 

1.80792

Hypothesis Testing Example

• Significance level = 5%

2 critical value = 5.991

• H

0

: (

• H

1

: (

,

) = (10000, 2)

,

) ≠ (10000, 2)

• ln LR = 2(-10034.660 + 10035.623) =1.207

• Accept H

0

Hypothesis Testing Example

• Significance level = 5%

2 critical value = 5.991

• H

0

: (

• H

1

: (

,

) = (10000, 1.7)

,

) ≠ (10000, 1.7)

• ln LR = 2(-10034.660 + 10045.975) =22.631

• Reject H

0

Confidence Region

• X% confidence region corresponds to the

1-X% level hypothesis test.

• The set of all parameters ( ,

) that fail to reject corresponding H

0

.

• For the 95% confidence region:

– (10000, 2.0) is in.

– (10000, 1.7) out.

Confidence Region

Outer Ring 95%, Inner Ring 50%

2.5

2.0

1.5

1.0

0.5

0.0

0 5000

Theta

10000 15000

Grouped Data

• Data grouped into four intervals

– 562 under 5000

– 181 between 5000 and 10000

– 134 between 10000 and 20000

– 123 over 20000

• Same data as before, only less information is given.

2.5

2.0

1.5

1.0

0.5

0.0

0

Confidence Region for

Grouped Data

Outer Ring 95%, Inner Ring 50%

5000

Theta

10000 15000

2.5

2.0

1.5

1.0

0.5

0.0

0

Confidence Region for

Ungrouped Data

Outer Ring 95%, Inner Ring 50%

5000

Theta

10000 15000

Estimation with Model Uncertainty

COTOR Challenge – November 2004

• COTOR published 250 claims

– Distributional form not revealed to participants

• Participants were challenged to estimate the cost of a $5M x $5M layer.

• Estimate confidence interval for pure premium

You want to fit a distribution to 250 Claims

• Knee jerk first reaction, plot a histogram.

Histogram of Cotor Data

250

200

150

100

50

0

0 1 2 3 4

Claim Amount

5 6 x 10

6

7

This will not do! Take logs

• And fit some standard distributions.

0.35

0.3

0.25

0.2

lcotor data lognormal gamma

Weibull

0.15

0.1

0.05

0

6 7 8 9 10 11 12

Log of Claim Amounts

13 14 15 16

Still looks skewed. Take double logs.

• And fit some standard distributions.

2.5

2

1.5

1

0.5

0

1.8

llcotor data

Lognormal

Gamma

Weibull

2 2.2

log log of Claim Amounts

2.4

2.6

2.8

Still looks skewed. Take triple logs.

• Still some skewness.

• Lognormal and gamma fits look somewhat better.

5

4

3

2 lllcotor data

Lognormal

Gamma

Normal

1

0

0.55

0.6

0.65

0.7

0.75

0.8

Triple log of Claim Amounts

0.85

0.9

0.95

1

Candidate #1

Quadruple lognormal

Distribution:

Log likelihood:

Lognormal

283.496

Domain:

Mean:

0 < y < Inf

0.738351

Variance: 0.006189

Estimate Std. Err. Parameter

Mu sigma

-0.30898

0.106252

0.00672

0.004766

Estimated covariance of parameter estimates:

mu sigma

Mu

Sigma

4.52E-05 1.31E-19

1.31E-19 2.27E-05

Candidate #2

Triple loggamma

Distribution:

Log likelihood:

Gamma

282.621

Domain:

Mean:

0 < y < Inf

0.738355

Variance: 0.00615

Estimate Std. Err. Parameter

A

B

88.6454

0.008329

7.91382

0.000746

Estimated covariance of parameter estimates: a b

A

B

62.6286 -0.00588

-0.00588 5.56E-07

Candidate #3

Triple lognormal

Distribution: Normal

Log likelihood: 279.461

Domain:

Mean:

-Inf < y < Inf

0.738355

Variance:

Parameter mu sigma

0.006285

Estimate Std. Err.

0.738355

0.079279

0.005014

0.003556

Estimated covariance of parameter estimates: mu sigma

mu

2.51E-05

sigma

-1.14E-19

-1.14E-19 1.26E-05

All three cdf’s are within confidence interval for the quadruple lognormal.

1

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0

0.55

0.6

lllcotor data

Lognormal

confidence bounds (Lognormal)

Gamma

Normal

0.65

0.7

0.75

0.8

Triple log of Claim Amounts

0.85

0.9

0.95

1

Elements of Solution

• Three candidate models

– Quadruple lognormal

– Triple loggamma

– Triple lognormal

• Parameter uncertainty within each model

• Construct a series of models consisting of

– One of the three models .

– Parameters within a broad confidence interval for each model .

– 7803 possible models

Steps in Solution

• Calculate likelihood (given the data) for each model.

• Use Bayes’ Theorem to calculate posterior probability for each model

– Each model has equal prior probability.

       

Steps in Solution

• Calculate layer pure premium for 5 x 5 layer for each model.

• Expected pure premium is the posterior probability weighted average of the model layer pure premiums.

• Second moment of pure premium is the posterior probability weighted average of the model layer pure premiums squared.

CDF of Layer Pure Premium

Probability that layer pure premium ≤ x equals

Sum of posterior probabilities for which the model layer pure premium is ≤ x

Numerical Results

Mean 6,430

Standard Deviation 3,370

Median 5,780

Range

Low at 2.5%

High at 97.5%

1,760

14,710

Histogram of

Predictive Pure Premium

Predictive Distribution of the Layer Pure Premium

0.08

0.06

0.04

0.02

0.16

0.14

0.12

0.10

0.00

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

Low End of Amount (000)

Example with Insurance Data

• Continue with Bayesian Estimation

• Liability insurance claim severity data

• Prior distributions derived from models based on individual insurer data

• Prior models reflect the maturity of claim data used in the estimation

Initial Insurer Models

• Selected 20 insurers

– Claim count in the thousands

• Fit mixed exponential distribution to the data of each insurer

• Initial fits had volatile tails

• Truncation issues

– Do small claims predict likelihood of large claims?

45,000

40,000

35,000

30,000

25,000

20,000

15,000

10,000

5,000

0

1,000

Initial Insurer Models

10,000 100,000

Loss Amount - x

1,000,000 10,000,000

Low Truncation Point

5,000

4,500

4,000

3,500

3,000

2,500

2,000

1,500

1,000

500

0

0.00

0.05

0.10

0.15

0.20

0.25

Probability That Loss is Over 5,000

0.30

0.35

0.40

5,000

4,500

4,000

3,500

3,000

2,500

2,000

1,500

1,000

500

0

0.00

High Truncation Point

0.01

0.02

0.03

0.04

Probability That Loss is Over 100,000

0.05

0.06

0.07

Selections Made

• Truncation point = $100,000

• Family of cdf’s that has “correct” behavior

– Admittedly the definition of “correct” is debatable, but

– The choices are transparent!

Selected Insurer Models

45,000

40,000

35,000

30,000

25,000

20,000

15,000

10,000

5,000

0

100,000 1,000,000

Loss Amount - x

10,000,000

6,000

Selected Insurer Models

5,000

4,000

3,000

2,000

1,000

0

0.00

0.01

0.01

0.02

0.02

0.03

0.03

Probability That Loss is Over 100,000

0.04

0.04

0.05

Each model consists of

1. The claim severity distribution for all claims settled within 1 year

2. The claim severity distribution for all claims settled within 2 years

3. The claim severity distribution for all claims settled within 3 years

4. The ultimate claim severity distribution for all claims

5. The ultimate limited average severity curve

Three Sample Insurers

Small, Medium and Large

• Each has three years of data

• Calculate likelihood functions

– Most recent year with #1 on prior slide

– 2 nd most recent year with #2 on prior slide

– 3 rd most recent year with #3 on prior slide

• Use Bayes theorem to calculate posterior probability of each model

Formulas for Posterior Probabilities

Model ( m ) Cell

Probabilities

P

, ,

F

  i

1

1

F

F

  i

Likelihood

( m )

Using Bayes’

Theorem

Number of claims l m

9 3   i

1 AY

1

P

 n

,

, ,

Posterior( ) l m

Results

Taken from paper.

1-3

1-3

1-3

1-3

1-3

1-3

1-3

1-3

1-3

1-2

1-2

1-2

1-2

1-2

1-2

1-2

1-2

1-2

1

1

1

1

Lags

1

1

1

1

1

Exhibit 1 – Small Insurer

Interval

Lower

Bound

Claim

Count

100,000 15

Layer Pure Premium

Prior Posterior $500K x $1M x

Model # Probability $500K

1 0.016406

763

$1M

541

200,000

300,000

2

1

2

3

0.041658

0.089063

911

1,153

645

682

400,000

500,000

750,000

1,000,000

1,500,000

2,000,000

2

0

0

0

0

0

4

5

6

7

8

9

10

11

0.130281

0.157593

0.110614

0.075702

0.053226

0.080525

0.104056

0.129925

1,224

1,281

1,390

1,494

1,587

1,849

2,069

2,417

796

912

978

1,040

1,095

1,328

1,523

1,828

100,000 40

200,000 10

300,000

400,000

1

0

500,000

750,000

1,000,000

1,500,000

2,000,000

2

0

0

2

0

12

13

14

15

16

17

18

19

20

0.010896

0.000007

0.000009

0.000011

0.000013

0.000014

0

0

0

2,598

2,788

3,004

3,202

3,382

3,543

4,058

4,663

5,354

1,916

1,922

2,124

2,309

2,477

2,628

3,211

3,784

4,440

Posterior Mean

Posterior Std. Dev.

1,572

463

1,113

385 100,000 76

200,000 26

300,000 11

400,000 3

500,000

750,000

1,000,000

1,500,000

2,000,000

0

0

8

0

0

Formulas for

Ultimate Layer Pure Premium

• Use #5 on model (3 rd previous) slide to calculate ultimate layer pure premium

20

Posterior Mean =

 m m

=1

Posterior Standard Deviation =

20 m 2  m m

=1

Results

Posterior Mean

Posterior Std. Dev.

Small Insurer Medium Insurer

Layer Pure Premium Layer Pure Premium

Large Insurer

Layer Pure Premium

$500K x $1M x $500K x $1M x $500K x $1M x

$500K $1M $500K $1M $500K $1M

1,572

463

1,113

385

1,344

278

909

245

1,360

234

966

188

• All insurers were simulated from same population.

• Posterior standard deviation decreases with insurer size.

Possible Extensions

• Obtain model for individual insurers

• Obtain data for insurer of interest

• Calculate likelihood, Pr{data|model}, for each insurer’s model.

• Use Bayes’ Theorem to calculate posterior probability of each model

• Calculate the statistic of choice using models and posterior probabilities

– e.g. Loss reserves

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