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MAXENTROPIC APPROACH TO DECOMPOUNDAGGREGATE RISK LOSSES
Henryk Gzyl Silvia [email protected] [email protected]
Erika Gomes [email protected]
Department of Business AdministrationUniversidad Carlos III de Madrid
IMEJune, 2015
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Research Question
Is it possible to infer thedistribution of the individual
severities from the aggregatedloss?
This is a topic of great interest to financial regulators and institutions, whooften only have access to highly aggregated data
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Literature Review
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Why this research?/Motivation
A risk analyst may be faced with the following problem:
He has obtained loss data collected during a year,but only contains the total number events andthe total loss for that year.
He suspects that there are different sources ofrisk, each occurring with a different frequency
He wants to identify the frequency with whicheach type of event occurs and if possible, theindividual losses at each risk event.
The knowledge of the individuals loss is required, because interesting details andvaluable information for risk management can be hidden and it is at this level
where the loss prevention or mitigation can be applied.
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What has been done?
Key featuresDisentangling:
Parametricmodel.
Assume thatdifferent risksareindependentand they haveuniquefeatures.
Involve visualcomparisonsand handlingsomeparameters bytrial and error.
Key featuresDecompounding:
Non-parametricmodel.
The resultingcurve is fX theunobservedmixture ofindividuallosses.
Only works inwell fittedfrequencydistributions
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Maximum Entropy Methods
First step: ψ(αk ) = E (e−αk S(N))
Analytical form
ψ(αk ) = G(φX (αk )) with αk = 1.5/k
Numerical form
ψ(αk ) =1
n
K∑k=1
eαk S(N) with αk = 1.5/k
where
φX (αk ): Laplace transform of X , αk ∈ R+ (not observed)
G(·): probability generation function of the frequencies (observed)
ψ(αk ): Laplace transform of the total losses (observed)
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Maximum Entropy Methods
H(f ) = −∫ 1
0fY (y)lnfY (y)dy
SME approach. Find the probability density on [0,1]∫ 1
0yαk f (y)dy = µ(αk ) with Y = e−S
SMEE approach: Extension of the SME approach when we assume that thedata has noise. ∫ 1
0yαk f (y)dy ∈ Ck = [ak , bk ] with Y = e−S
where µ = ψ(αk )−P(N=0)1−P(N=0)
These methods consist in to find the probability measure which best representthe current state of knowledge which is the one with the largest information
theoretical entropy.
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DecompoundingIt is not be always possible to observe frequency and severity separately. At this pointwe may want to know the distribution of individual losses, and the maxent methods canhelp us to find it.
E(e−αk S ) = G(φ(αk )) =1
n
K∑k=1
eαk S(N) ,
where k = 1, . . . ,K .
Example:
G(z) =∞∑
n=0znP(N = n) = e`(z−1), where N ∼ Poiss(`).
φ(αk ) = E(e−αk Xi ), laplace transform of the individual loss Xi .
ψ(αk ) = 1n
∑Kk=1 eαk S(N) = e`(φ(αk )−1), laplace transform of the total losses
Then the input is,
φ(αk ) = 1`
ln(ψ(αk )) + 1, where |φ(αk )| < 1
` = `1 + `2 + `3 + ...
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Disentangling
Suppose that we have frequency data that does not distinguish between subpopulationsof risk sources, then it is necessary
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DisentanglingPanjer Recursion: Class (a,b,0)
Let N be a random variable taking positive integer values, and write pk = P(N = k)for k ∈ N. We shall say that N is in the class (a, b, 0) if there exist constants a, b suchthat
pk/pk−1 = a + b/k; for k = 1, 2, 3, ...
Distribution a b p0
Poisson 0 λ e−λ
Binomial − p1−p
(n + 1) p1−p
(1− p)n
Neg. Binomial β1+β
(r − 1) β1+β
(1 + β)−r
Geometric β1+β
0 (1 + β)−r
k · r(k) = ak + b; r(k) = pk/pk−1
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Simulation details
To test the methodology we consider different combinations of fre-quencies and severity losses.
We do several cases using Poisson and Negative Binomial ascounting distribution because they are rather common ininsurance and Operational Risk models.
In all numerical examples we measure the quality of thereconstructions by several distances between the reconstructeddensity and the histogram as well as between thereconstructed density and the equivalent density mixture.
We consider samples between 200 and 500 data points.
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Results (Disentangling)Case 1. Poisson- LogNormal distribution
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Results (Disentangling)Case 1. Poisson- LogNormal distribution
k ·r(k) = ak+b; r(k) = pk/pk−1
Distribution a b
Poisson 0 λBinomial − p
1−p (n + 1) p1−p
Neg. Binomial β1+β (r − 1) β
1+β
Geometric β1+β 0
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Results (Disentangling)Case 1. Poisson- LogNormal distribution
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Results (Disentangling)Case 1. Poisson- LogNormal distribution
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Results (Decompounding)Case 1. Poisson- LogNormal distribution
The decompounding procedure only works over well fittedfrequency models
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Results (decompounding)Case 1. Poisson- LogNormal distribution
In all numerical examples we measure the quality of the reconstructions by severaldistances between the reconstructed density and the histogram as well as between thereconstructed density and the equivalent density mixture.
MAE =1
N
N∑i=1
|Fi − Fn,i |, RMSE =
√√√√ 1
N
N∑i=1
(Fi − Fn,i )2
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Conclusions
Key featuresDisentangling:
Parametricmodel.
Assume thatdifferent risksareindependentand they haveuniquefeatures.
Involve visualcomparisonsand handlingsomeparameters bytrial and error.
Key featuresDecompounding:
Non-parametricmodel.
The resultingcurve is fX theunobservedmixture ofindividuallosses.
Only works inwell fittedfrequencydistributions
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Conclusions
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Conclusions
We presented a two stage procedure to determine the probabilitydensity of individual losses corresponding to an observed aggregateloss.
Determine a disentangling model that is good enough dependsof the particular application, therefore previous knowledgeabout the data might be useful for the analysis. For example,in operational risk, the Poisson as well as the negativeBinomial models are adequate to describe the frequency ofthe losses.
Decompounding serves as a method to verify the quality ofthe first step, if the frequency model is to far from the real,the decompounding method would give bad results.
Decompounding method shows good results for any well fittedparametric distribution.
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QUESTIONS, COMMENTS
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