Theoretical and Practical Aspects of Parameter and Model

36th ANNUAL GIRO CONVENTION Theoretical and Practical Aspects of Parameter and Model UncertaintyEdinburgh October 2009

Dimitris PapachristouAON BENFIELD

Topics

Sources of parameter uncertaintyDiversifiable and non diversifiable parameter riskEffect of model/parameter uncertainty on a single risk and on a portfolio of risksMethods of estimating the parameter uncertainty: analytical, Monte Carlo, asymptotic approximationsPitfalls and commonly used methods Model uncertainty and common actuarial distributionsApplications: frequency of aviation losses, risk transfer assessment

Sources of Uncertainty of Some Risk Quantity

Process Risk: due to the stochastic nature of the risk given a statistical model and its parametersParameter Uncertainty: uncertainty in the values of the parameters of the statistical modelModel Uncertainty: uncertainty in the choice of the statistical model

Types of Parameter and Model Uncertainty

The nature of the parameter and model risk is similar

statistical uncertainty due to the limited amount of datauncertainty because of heterogeneity in the portfolio.

For example a portfolio of motor policyholders where there is a mix of drivers of different skills

uncertainty because of change in the parameters over timeParameter uncertainty because of errors in the data or uncertainties in our estimates of the losses,

although it is an important source of uncertainty, it is not examined in this paper.

Diversifiable and non diversifiable parameter risk

Three Experiments - Nature of Parameter Uncertainty

1st Experiment: Fair coin with known p=0.52nd Experiment: Several coins, each has a different p and p~U(0,1)3rd Experiment: One coin, p~U(0,1), same coin is used for several throws

Three Experiments Nature of Parameter Uncertainty

1st Experiment: Fair coin with p=0.5. Insurance analogy:

we have sufficient information to assess the parameters of the risk with reasonable accuracy, or simply a decision has been made to ignore parameter risk

2nd Experiment: Several coins, each set of trials has a different p and p~U(0,1). Insurance analogy:

a heterogeneous portfolio of policies. e.g. a portfolio of motor policies where the drivers have varying driving skills and for each of the drivers the risk parameter isdifferent

3rd Experiment: One coin, p~U(0,1). Insurance analogy:limited information to assess the parameters of a single type ofrisk

Three Experiments Nature of Parameter Uncertainty

1st Experiment: Fair coin with p=0.5

2nd Experiment: Several coins, each set of trials has a different p and p~U(0,1)

3rd Experiment: One coin, p~U(0,1), several throwsIn the third experiment I would prefer to look at the probability of heads as a random variable

which is a function of the random variable , with .

knk ppkn

kNP −−⎟⎟⎠

⎞⎜⎜⎝

⎛== )1(][

∫ −−⎟⎟⎠

⎞⎜⎜⎝

⎛==

1

0

)1(][ dpppkn

kNP knk

knk ppkn

kNP −−⎟⎟⎠

⎞⎜⎜⎝

⎛== )1(][

p )1,0(~ Up

Parameter uncertainty due to limited amount of data

The process risk can be diversified over time or over a portfolio of similar and independent policies.

the degree of diversification may be limited by the time horizon of the risk taker or the number of policies in the portfolio

The parameter and model risk for a single type of policy can not always be diversified over time. The model and its parameters are fixed but unknown.

The belief about the value of the parameters could take the form of a prior distribution or combined with some available data the form of a posterior distribution. In practice we do not have a purely homogenous portfolio (single type) of policiesAs time passes more data is collected and the parameter estimates and parameter distributions are updated


Summary descriptions of the risk such as the mean make more sense when the risk can be diversified. The mean of the process risk distribution shows the expected loss over time or over a very large portfolio of identical and independent policies. The mean of the prior or the posterior distribution for the riskparameters can always be calculated but its meaning may not necessarily have an intuitive practical interpretation. When insurance practitioners, especially practitioners not mathematically trained, refer to the “1 in 10 years event” it is doubtful whether they refer to both the space of the possible loss outcomes and the space of the possible parameter and model values.


The discussion touches on the old debate between “frequentists” and “Bayesians”I accept both philosophical interpretations of probability and their applications, but I distinguish in risk applications between probability as an opinion and probability as frequency of experiment.

Comparison: Mixture Distribution – Fixed Known Parameter – Fixed but Unknown Parameter

First Case: Fixed and known ParameterLet assume that is a random variable following an distribution where the parameter is fixed and known.

A p-percentile is given by

)(λlExponentia λ

0,,1);( >−= − xexF x λλ λ

λλ )1ln();( pph −−

=

X


Second Case: Heterogeneous Portfolio – Mixture DistributionNow let assume that and is a random variable which follows a distribution. The unconditional distribution function of is

A p-percentile is given by

)(~| λλ lExponentiaX λ

),( baGamma

X0,,,1

)(),;(

0 0

1

>⎟⎠⎞

⎜⎝⎛

+−=⋅

Γ= ∫ ∫

∞ −−− bax

xbbdyd

aebebaxF

ax baay λλλ

λλ

( )b

pbbaph a −−

= /11),;(


Third Case: Single Policy with Unknown ParameterHere the percentile is a random variable because is a random variable which follows a distribution. Although the mean of this random variable does not have a clear intuitive interpretation, someone could estimate the mean of

λλ )1ln();( pph −−

=

λ),( baGamma

λλ )1ln();( pph −−

=

[ ] ( )∫∞ −−

−−−

=Γ

−−=⎥⎦

⎤⎢⎣⎡ −−

=0

1

1)1ln(

)()1ln()1ln();(

abpd

aebppEphE

baa

λλλλ

λλ

1>a


Numerical ExampleIf when it is fixed and known If follows a distribution when assumed to be a r.v. with

First case: =3.00,

Second case: =15.66 and

Third case: =27.23 The results differ significantly in these three casesThe ordering is not always the same

1=λλ )1.1,1.1( == baGamma

1][ =λE

95.0=p

λλ )1ln();( pph −−

=

( )b

pbbaph a −−

= /11),;(

[ ])1(

)1ln();(−−−

=ab

pphE λ

Simulation: Mixture Distribution - Fixed but Unknown Parameter

Simulation Method 1:A parameter is simulated and then a loss is simulated, given the simulated parameter. This is repeated many times. The 95th percentile of the empirical simulated distribution is the estimate for the 95th percentile.

Simulation Method 2:A parameter is simulated and then the 95th percentile of the distribution given the simulated parameter is calculated. Repeating this process a number of times an estimate of the distribution of the 95th percentile can be obtained

The first method is more appropriate for the case where we have a heterogeneous portfolio of policies.

The second method is more appropriate when we have a single type of risk and the parameter is considered to be a random variable.

the first method tends to be more common irrespective of whether one policy or a portfolio of policies with different parameters is examined.

Portfolio of PoliciesHeterogeneous portfolio

e.g frequency of motor accidents follows and is different for each driver and follows

Generally holds

For a heterogeneous portfolio the process and parameter risk can be diversified by increasing the number of independent policies

In practice there may be some dependency between the policies

but this will not be due to the different risk parameters

)(λP

),( baGammaλ

]]|[[][ θXEEXE = ]]|[[]]|[[][ θθ XVEXEVXV +=

Portfolio of PoliciesHomogeneous portfolio

e.g. lives of identical risk whose probability of death depends on a life table. The parameters of the table are subject to estimation error

The process risk can be diversified by increasing the number of policies,

But the parameter risk can not be diversified by increasing the number of identical policies

my preferred treatment of quantities such as percentiles of the loss to the portfolio is to consider these quantities as random variables and present their distribution

However, some may want to treat the space of the values of the parameters in the same way they would treat the space of the loss outcomes

Alternatively we could consider groups of large portfolios each with each own parameter

Portfolio of PoliciesResult 1 : If are independent and identically distributed random variables then it can be shown that

This is an interesting result on its own when correlation is modelled through a common parameterResult 2: If are independent and identically distributed random variables then

The standard deviation per policy does not go to 0 as the number of policies goes to infinity. The parameter risk can not be diversified

∞=Θ 1}|{ jjX

]]|[[),( Θ= iji XEVXXCov

),()1(][1

jii

n

ii XXCovnnXVnXV −⋅+⋅=⎥⎦

⎤⎢⎣

⎡∑=

]]|[[),(1 Θ=→⎥⎦

⎤⎢⎣

⎡∑=

iji

n

ii

XEVXXCovn

XV

∞=Θ 1}|{ jjX

∞→n

Portfolio of PoliciesNumerical example

e.g frequency of motor accidents follows and

)(λP

)12,4(~ == baGammaλ

number of policies in the portfolio

same lamda for each policy

different lamda for different policies

1 0.6009 0.600910 0.2472 0.1900

100 0.1764 0.06011,000 0.1677 0.0190

10,000 0.1668 0.0060infinite number 0.1667 0.0000

standard deviation per policy

Application 1: Risk Transfer

Probability of reinsurer’s negative resultNo specified threshold sometimes 10% probability of 10% Loss(PV of Losses – PV of Premiums)/PV PremiumsIncludes Profit Commissions and Additional PremiumsOnly cash-flows between reinsurer and reinsured considered

Expected Reinsurer’s Deficit (ERD) the conditional expectation of Reinsurer’s Loss is consideredno standard threshold - usually around 3%

Wang TransformationVan Slykes – Kreps ApproachOther

Application 1: Risk TransferReport to the Casualty Actuarial Task Force of the NAIC

about 37% of the respondents in their survey takes into account parameter uncertainty when calculating risk transfer. emphasises the importance of parameter uncertainty in the risk transfer calculations

Common method: simulate the parameters and then simulating losses using the simulated parameters. for each simulation the risk transfer measurement is calculated and some statistics are calculated. Casualty Actuarial Task Force uses implicitly in one of the example of their report.

The assessment of risk transfer is made on an individual contract basis. This means that the parameter risk can not be diversified. whatever risk transfer measure is used, it is a random variable whose value depends on the unknown parameters. E.g. the ERD is not a single number as it is commonly perceived to be, but a random variable with a distribution which depends on the distribution of the parameters.


Example ContractTerm 3 yearsAnnual Limit 15Term Limit 30Annual Deductible 30Premium 5Reinsurer’s margin 30% of premiumAdditional Premium 25% of annual lossesProfit Commission Maximum of (Premium – Margin –Losses, 0)


Losses to the contract are assumed to follow a Pareto(c=10 and a=1.2). The parameter is assumed to have been estimated based on 10 losses For simplicity the discount rate is assumed to be 0%.If the simulations are done in the common way then the ERD is approximately 3.5%.

Application 1: Risk TransferThe process risk can be diversified over time. The parameter risk in this case can not be diversified.

The ERD is not a single number but a random variable.

There is a non negligible probability that the ERD takes values significantly higher than the 3.5% value which is estimated with the commonly used method of calculation.

0%

10%

20%

30%

40%

50%

60%

70%

80%

90%

100%

0% 1% 2% 3% 4% 5% 6% 7% 8% 9% 10% 11% 12% 13% 14% 15% 16%ERD

Cum

ulat

ive

Dis

trib

utio

n

Application 1: Risk TransferSomeone could use the average of the ERD distribution as a measure of risk transfer which in this case is 3.7% However, a summary number does not necessarily adequately present the amount of risk.

It would be similar to trying to base decisions about the value of an excess of loss reinsurance programme on the expected loss rather than on the whole distribution of the loss.

I have seen reinsurance transactions with a genuine risk transfer motivation and a significant loss uncertainty such that several reinsurers did not accept the risk. parameter uncertainty being one of the main sources of uncertainty.

However, the usual actuarial simulation methods of averaging over the space of parameters estimated risk transfer measures just above the usual benchmarks. The averaging over the space of parameter values failed to show one of the main drivers of risk and one of the primary motivations for the transaction.

Likelihood function and estimation

Bias in MLE estimatesExample: Pareto

MLE is asymptotically unbiased, but unfortunately our world is not asymptoticConsider

In practice it is usually capped

The lower the value of parameter , the “fatter” the tailDataMLE estimate

0,01)( ><<⎟⎠⎞

⎜⎝⎛−= axc

xcxF

a

nxx ,...,1

∑=

⎟⎠⎞

⎜⎝⎛

= n

i

i

cx

na

1ln

ˆ

a


If the true parameter is it can be shown that the distribution of the MLE estimator has p.d.f.

It can be shown that

…which on average is higher than . (Lighter tail)Also it is more likely the estimate to be higher than

0a

( ))(

)( 10

0

naeanaf n

ana

n

Γ⋅

= +

⎟⎠⎞

⎜⎝⎛−

1][ 0

−⋅

=n

anaE

0a0a


Assume true parameter is 1.2Both median and mean of estimator are higher than 1.2

0

0.5

1

1.5

2

2.5

3

3.5

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3a

Prob

abili

ty D

ensi

ty F

unct

ion

10 points20 points100 points

A “Paradox”Consider the p x 100-th percentileIf a is our estimate for the parameter, then the p100-th percentile is

It can be shown that under certain conditions

“Paradox”: Although on average our estimate of the parameter is higher (lighter tail) than its true value, on average the percentiles are higher than their true valueHowever, it is more likely to underestimate the percentile (as it is more likely to overestimate the value of the parameter)

apcY1

)1(−

−⋅=

n

panancYE ⎟⎟

⎠

⎞⎜⎜⎝

⎛−+⋅

⋅=

)1ln(][

0

0

Distribution of the 95th percentileAssuming c=10, a=1.2 and n=10

Unbiased parameter estimator does not necessarily imply unbiased percentile estimator

0%

10%

20%

30%

40%

50%

60%

70%

80%

90%

100%

0 200 400 600 800 1000 120095th percentile

Cum

ulat

ive

Dis

butio

n

true 95th percentile

average 95th percentile

Likelihood Function

Same example as before with Pareto sampleThe likelihood function is

Which has the form of a

∑∝

∑⋅== ==

−+−

=

+−∏n

ii

n

ii xa

nxa

nn

i

ai

n eaeaxaaL 11

)ln()ln()1(

1

)1()(

⎟⎠

⎞⎜⎝

⎛+ ∑

=

n

iixnGamma

1)ln(,1

)(

))ln(()(

1

1

)ln(1

a

xeaaf

nn

ii

xan

n

ii

Γ

∑

=

+

=

−

∑=

Likelihood Function - BootstrapCompareIf the true parameter is it was shown that the distribution of the MLE estimator has p.d.f.

The likelihood function is

The two are different. The first is an inverse gamma distribution, the second a gamma distribution. In the first case the parameter is given, in the second the sample is given.

0a

( ))(

)( 10

0

naeanaf n

ana

n

Γ⋅

= +

⎟⎠⎞

⎜⎝⎛−

)(

))ln(()(

1

1

)ln(1

a

xeaaf

nn

ii

xan

n

ii

Γ

∑

=

+

=

−

∑=

Likelihood Function - BootstrapCommon practice

We estimate the parameterWe assume it is the correct one and we simulate data samples based on the estimated parameterFor each sample we estimate the parameterWe construct the distribution of the parameter

This is not strictly correct because we assume that the parameter is given, while what is given is the sample

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4Pareto Parameter a

Prob

abili

ty D

ensi

ty F

unct

ion

estimator distribution (1st case)likelihood function (2nd case)

Numerical ExampleAssume Pareto(c=1,a=1.2) and simulated data

Likelihood estimate is a=1.6. (if true a=1.2, there is a probability of around 20% that likelihood estimate will be greater than a=1.6)The true 95th percentile is 12.139.

The 95th percentile based on the maximum likelihood estimate for is 6.503

The expected value of the 95th percentile when allowing for parameter uncertainty is

Allowing for parameter uncertainty does not necessarily results in percentile estimates which are close to the true ones

1.215 1.383 2.203 1.360 1.171 1.304 5.511 1.604 4.409 1.725

=−−

2.11

)95.01(

=−−

6.11

)95.01(

851.8)95.01(][1

95.0 ≅⎥⎦

⎤⎢⎣

⎡−=

−aEYE

Calculation and Estimation Methods of the Parameter Distribution

Calculation of Parameter Distribution

Exact distribution has a “nice” known analytical formNormal approximationMonte Carlo methods

Asymptotic v. Actual Distribution of Weibull Parameter Estimator based on 11 data points

Asymptotic v. Actual Distribution of Weibull Parameter Estimators

11 points

20 points

50 points

Asymptotic v. Actual Distribution of Weibull Parameter Estimators

Actual distribution does not look like its normal approximation when data sparseNormal Distribution can produce negative values for parameters which are positive by definitionIgnore negative values?

Convenient, but not always a good approximation!

Calculation of Parameter Distribution

The Pareto examples discussed earlier lead into nice analytical formulae for the likelihood which had a gamma form.

However, this is not generally the case. For example Weibullhas likelihood function

Which does not have a recognisable standard form

)(

))ln(()(

1

1

)ln(1

a

xeaaf

nn

ii

xan

n

ii

Γ

∑

=

+

=

−

∑=

0,,,)( 1 >⋅⋅= −− γγγγ cxexcxf cx

∑∑=

∑= ===

−−

=

−−

∏n

ii

n

ii

n

ii xxc

nnn

ii

xcnn eecxeccL 111

)ln()1(

1

1),(γ

γγγ

γγγ

Calculation of Parameter DistributionMonte Carlo Methods

Monte Carlo Statistical Methods such as the Gibbs sampler can be of great help in these situations. These methods have found applications in Bayesian Statistics which are briefly mentioned later, but they can also be used in the classical case. The details of the Monte Carlo methods are not discussed here, but can be found in Robert, C. P. & Casella , G. (2004). A shorter description which is also more relevant to actuarial work can be found in Scollnik, D. P. M. (2000).Exact method used depends on the problem, here the generic case is shown

Calculation of Parameter DistributionMonte Carlo Methods – Gibbs Sampling

Let say we want to simulate from the joint density Gibbs Sampling

Initial values are arbitrarily chosenThen s are simulated from the conditional distributions

This is repeated many times the first few simulated values are usually ignored If the conditional distributions can not be recognised, then a generic sampling algorithm usually based on variations of the Metropolisalgorithm is used

),...,,( 21 kf θθθ

)0()0(2

)0(1 ,...,, kθθθ

θ

),...,|(~..

)...,,|(~

),...,|(~

)1(1

)1(1

)1(

)0()0(3

)1(12

)1(2

)0()0(21

)1(1

−kkk

k

k

f

f

f

θθθθ

θθθθθ

θθθθ

Calculation of Parameter DistributionMonte Carlo Methods – Gibbs Sampling

Example: For the Weibull distribution we have

has the form of a but does not have a standard easily recognisable formRandom numbers from could be simulated using some version of the Metropolis algorithm

∑∝ =

−n

iixc

neccf 1)|(γ

γ

∑∑∝ ==

−n

ii

n

ii xcx

n eecf 11

)ln(

)|(γγ

γγ)|( γcf ⎟

⎠

⎞⎜⎝

⎛+ ∑

=

n

iixnGamma

1,1 γ

)|( cf γ

)|( cf γ

The Metropolis algorithmUsed to simulate losses (under some conditions) from a distribution which does not have a standard recognisable formThere are many variations of the algorithm

The appropriate variation depends on the nature of the problem and the form of the distribution

Generate A is sufficiently largeGenerate

If then

Else

),0(~1 AUniformU

)1,0(~2 UniformU

⎟⎟⎠

⎞⎜⎜⎝

⎛<

−

1,)()(

min1

12

tafUf

U 1Uat =

1−= tt aa

Temporal Parameter Uncertainty

Temporal Parameter UncertaintyChanges may be gradual or suddenEffect of change may be different on different risk covers, may affect severity or frequency or bothChanges may be estimated using external information or from the data

A good measure of exposure may provide a better indication of risk changes compared to what can be extracted from some sparse data some special features may not be described sufficiently by any external information, then the loss data may be able to reveal some trends

When data used, some non parametric methods may be useful in extracting trends, see laterIt is often the case that trends are hidden in randomness and can not be identified or measured easily

Temporal Parameter Uncertainty – Fooled by Randomness

Assume frequency follows Poisson( ). Poisson parameter changes in a compound way by Experiment: investigate the behaviour of the estimatorsThe log-likelihood function is

For =3, g=10% and number of years n=10

gk += 1

( ) )ln()1()ln(...1),(11

1 kninkkkln

ii

n

ii

n ⎟⎠

⎞⎜⎝

⎛−+⎟

⎠

⎞⎜⎝

⎛++++−= ∑∑

==

− λλλ

λ

λ

Temporal Parameter Uncertainty – Fooled by Randomness

For =3, g=10% and number of years n=10, based on 5,000 sim

Random fluctuations in the simulated number of losses can be interpreted as growth trends

λ

-10%

-5%

0%

5%

10%

15%

20%

25%

30%

35%

40%

0 1 2 3 4 5 6 7 8

parameter lamda

grow

th g

Effect of “Unusually” Large Losses

“Unusually” Large Losses and Rules of Thumb

Large losses are treated separately. What is a large loss? Rules of thumb, e.g. largest loss is 3 times larger than the second largestAssume Pareto a=1.2 and follow the above rule of thumb

There is significant difference in the distribution of both the parameter and the 95th percentile if we exclude an “unusually” high loss.

0%

10%

20%

30%

40%

50%

60%

70%

80%

90%

100%

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4Pareto Parameter

Cum

ulat

ive

Dis

trib

utio

n

using all dataexcluding 'unusual' losses

0%

10%

20%

30%

40%

50%

60%

70%

80%

90%

100%

0 100 200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500

95th percentileC

umul

ativ

e D

istr

ibut

ion

using all dataexcluding 'unusual' losses

“Unusually” Large Losses and Rules of Thumb

If is the kth upper order statistics, i.e. is the maximum of a sample of n and is the minimum of a sample of size n, then it can be shown that

Apply to Pareto

For a=1.2 and there is a probability of 26.75% that the largest loss will be more than 3 times higher than the second largest loss. a large loss should not be considered to be “unusual” without careful examination.

nkX , nX ,1nnX ,

11

1,1, ,)(1

)(1)|Pr[ ++

++ >⎟⎟⎠

⎞⎜⎜⎝

⎛−−

==> k

k

kknknk xy

xFyFxXyX

1,1]|Pr[

,2

,2,2,2,2,1 >⎟

⎠⎞

⎜⎝⎛=

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅==⋅> λ

λλ

λa

a

k

a

nnnnn

xc

xc

xXxX

3=λ

Model Uncertainty

Model UncertaintyIn the previous slides the model (distribution) was assumed known and the parameter was estimatedThe model (distribution) could also be estimated by the dataLikelihood function could also be used

Different distributions has different number of parameters and several criteria have been proposed for comparing different modelsExample criterion is the Schwartz Bayesian Criterion (SBC) which is

Ln(likelihood) –

Where r is the number of parameters and n is the sample size

⎟⎠⎞

⎜⎝⎛⋅π2

ln nr

Model Uncertainty: ExperimentAssume that the true distribution is Pareto(c=10,a=1.2)Random samples are generated and the following distributions are fitted: Pareto, Exponential, Weibull, Gamma, LogNormal and LogGammaThis is repeated 10,000 times

Model Uncertainty: Experiment

The distributions of the 95th percentile are:

0%

10%

20%

30%

40%

50%

60%

70%

80%

90%

100%

0 100 200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 150095th percentile

Cum

ulat

ive

Dis

butio

n

only Paretotrue 95th percentileselection of distributions

true 95th percentile

Model Uncertainty: ExperimentFor a sample size of 10

the true distribution (Pareto) is only chosen 27.1% of the time The LogNormal distribution is chosen more often that the true underlying model

when a Pareto model is assumed the 95th percentile is underestimated about 55% of the time and when a selection of distributions is used the 95th percentile isunderestimated about 65% of the time In social sciences the use of distributions is often justified by data and general reasoning.

E.g. it has been argued that the distribution of the size of cities, the size of the earthquakes or the income of people follows a Pareto type distribution. There have not been similar studies for the loss distributions in insurance.

Results are subject to simulation error

Application 2: Frequency of Aviation LossesNumber of losses which include at least 1 fatality from a subset of commercial airlines

Application 2: Frequency of Aviation Lossesa non parametric is helpful in exploring possible temporal trends in the data. the number of losses in year t are assumed to follow a

distribution with, where is the exposure for year t and a smoothing spline.

)( tPoisson μ )ln()()ln( tt Ets +=μ tE)(ts

5 10 15

-2-1

01

y1

s(y1

)

5 10 15

-2-1

01

2

y1

s(y1

, df =

1)

Application 2: Frequency of Aviation Lossescontinuous reduction in the expected number of losses, anova analysis is inconclusive as to whether the trend has accelerated after 2001 (WTC).

1995 2000 2005

24

68

10

year

Num

ber o

f Los

ses

Red: smoothing splineBlue: straight line

Application 2: Frequency of Aviation LossesAssume ork can be thought as k=1+g , where g is the annual rate of change in the Poisson parameter.

and are to be estimated. The likelihood function is

Gibbs sampler

is not in the form of a standard distribution and the Metropolis algorithm can be used

)ln()ln( tt Ebta ++=μ tt

tbta

t EkEe ⋅⋅== + λμ

( ) ( )∏∏=

−−

=

−− ⋅∑

⋅∑

== ==

−−

n

t

tnngEn

tt

ntt

gE t

n

tt

n

t

tt

ttt gengEegl

1

1

1

1 11

11

/),( λλλλ

λ

λ g

⎟⎠

⎞⎜⎝

⎛+∝ ∑∑

==

n

tt

n

tt EnGammagl

11,1)|(λ

)|( λgl

Application 2: Frequency of Aviation Losses

0%

10%

20%

30%

40%

50%

60%

70%

80%

90%

100%

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5Poisson Parameter

Cum

ulat

ive

Dis

trib

utio

n

Application 2: Frequency of Aviation Losses

Growth parameter negatively correlated with initial Poisson parameterEstimate of growth parameter is -11.7% and its 5th

percentile is -15.8% and its 95th percentile is -7.1%The estimate for the number of losses is 1.87 per year with its 5th and 95th percentile 0.8 and 3.3 respectively The choice of the parameter needs to take into account other qualitative information related to the airlines industry

Observations and Comments (1)Whether parameter risk can be diversified or not depends on the source of the parameter uncertaintyThe method of estimation/simulation also depends on the source of parameter uncertainty and whether it can be diversified or notUsing the wrong method could result in underestimation and poor understanding of the risk The correct distribution of the parameter estimator is given by the likelihood function, given the sampleThe distribution of the estimator given its true value can be used for investigations and for validation of methodsAllowing for parameter uncertainty does not necessarily results in percentile estimates which are close to the true ones

Observations and Comments (2)The Normal approximation to the parameter distribution is convenient but not always goodMonte Carlo methods can be used insteadRandom fluctuations in the simulated number of losses can be interpreted as growth trendsa large loss should not be considered to be “unusual” without careful examination of the properties of its distribution

The views expressed in this presentation are my personal views and in particular they should not necessarily be regarded as being those of my employer

ReferencesAMERICAN ACADEMY OF ACTUARIES – COMMITTEE ON PROPERTY AND LIABILITY FINANCIAL REPORTING (2005) Risk Transfer in P&C Reinsurance – Report to the Casualty Actuarial Task Force of the NAIC.BINMORE, K. (2009) Rational Decisions, Princeton University PressBUHLMANN, H. & GISLER, A. (2005) A Course in Credibility Theory and its Applications, SpringerCAIRNS, A. J. G. (2000) A Discussion of Parameter and Model Uncertainty in Insurance, Insurance Mathematics and Economics 27, p.313-330 CASELLA, G. & GEORGE, I. E. (1992) Explaining the Gibbs Sampler, The American Statistician, Vol. 46, No 3 EMBRECTHS, P., KLUPPELBERG, C. & MIKOSCH, T. (1997) Modelling Extremal Events, SpringerFASB (1992) FAS 113 Accounting and Reporting for Reinsurance of Short-Duration and Long Duration ContractsGAZZANIGA, M. (2008) Human: The Science behind What Makes us Unique, EccoGILLIES, D. (2000) Philosophical Theories of Probability, RoutledgeHALL, P & TAJVIDI, N (2000) Nonparametric Analysis of Temporal Trend when Fitting Parametric Models to Extreme –Value Data, Statistical Science, Vol. 15, No. 2, 153 - 167KLUGMAN, S., PANJER, H. & WILLMOT, G. (1998) Loss Models, WileyMAJOR, J. (1999) Taking Uncertainty into Account: Bias Issues Arising from Parameter Uncertainty in Risk Models, Casualty Actuarial Society Forum, Summer 1999MATA, A. (2000) Parameter Uncertainty for Extreme Value Distributions, GIRO ConventionMILLNS, R. & WESTON, R. (2005) Parameter Uncertainty and Capital Modelling, GIRO ConventionPRESS, W. H., FLANNERY, B. P., TEUKOLSKY, S. A. & VETTERLING, W. T. (1994) Numerical Recipes in Pascal, Cambridge University PressROBERT, C.P. & CASELLA, G. (2004) Monte Carlo Statistical Methods, SpringerROOTZEN, H. & TAJVIDI, N. (1995) Extreme Value Statistics and Wind Storm Losses: A Case Study, Scandinavian Actuarial JournalSCHMOCK, U. (1999) Estimating the Value of the WINCAT Coupons of the Winterthur Insurance Convertible Bond – A Study of the Model Risk, ASTIN Bulletin, Vol. 29, No 1, pp 101-163SCOLLNIK, D.P.M. (2000) Actuarial Modelling with MCMC and Bugs, North American Actuarial Journal, Vol. 5, No. 2, pp 96 - 124SORNETTE, D (2000) Critical Phenomena in Natural Sciences, SpringerTIERNEY, L. (1994) Markov Chain for Exploring Posterior Distributions, The Annals of Statistics, Vol. 22, No 4. pp 1701 - 1728WACEK, M (2005) Parameter Uncertainty in Loss Ratio Distributions and its Implications, Casualty Actuarial Society Forum, Fall 2005

Documents

Theoretical and Practical Aspects of Parameter and Model