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Point Process Techniques in Non-Life Insurance Models

Thomas Mikosch

University of Copenhagen

1Conference in Honor of Jan Grandell, Stockholm, June 13, 2008

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What is a point process?

• Consider the arrivals of claims in a portfolio:

0 < T1 < T2 < · · · .

• Count the number of arrivals in a set A

NT(A) = #{i ≥ 1 : Ti ∈ A} ,

finite if A is bounded. This defines a point process

NT = (NT(A))A∈E on the state space E = (0, ∞). • It is a random counting measure.

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The principle of marking

Marking with the claim sizes.

• If the Ti’s constitute a point process, so do the pairs (Ti, Xi) where Xi > 0 are the claim sizes corresponding to the claim arriving at Ti: for sets A, B ⊂ E,

NT,X(A × B) = #{i ≥ 1 : Ti ∈ A , Xi ∈ B}

The state space, where the points (Ti, Xi) live, is E = (0, ∞) × (0, ∞).

• The process NT,X is a marked point process: Xi is the mark

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Marking with the claim sizes and reporting delays.

• Claims are not reported at the times when they incur. They are reported with some positive delay in reporting, Di > 0. • The points (Ti, Xi, Di) constitute a marked point process

NT,X,D(A × B × C) = #{i ≥ 1 : Ti ∈ A, Xi ∈ B, Di ∈ C} .

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Marking with the claim sizes, reporting delays and settlement times.

• Claims are not settled when they are reported. It usually takes some settlement time Si > 0 before the amount of the claim size Xi is paid to the insured.

• The points (Ti, Xi, Di, Si) constitute a marked point process

NT,X,D,S(A × B × C × D)

= #{i ≥ 1 : Ti ∈ A, Xi ∈ B, Di ∈ C, Si ∈ D} .

• The state space is

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• One could also mark with a (random) function describing the process of settlement. Then the state space becomes more interesting.

• The notion of point process is too general to be useful.

• One needs assumptions on the distribution of the point process. • One such assumption is that the process be Poisson.

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Who invented the Poisson process?

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Who invented the Poisson process?

Filip Lundberg (1903)

according to A.N. Shiryaev

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What is a Poisson process (Poisson random measure)?

• It is a point process on a state space E. (Random measure) • There exists a Radon2 mean measure µ on (E, E) such that

1. N (A) is Poisson(µ(A)) distributed

2. N (A1), . . . , N (Am) are independent for A1, . . . , Am disjoint. • We say that N is PRM(µ) on E.

Example. The claim arrivals 0 < T1 < T2 < · · · constitute a

homogeneous Poisson process with intensity λ > 0 if

EN (a, b] = λ |(a, b]| = λ (b − a) .

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Marked Poisson processes are Poisson processes

• If Ti constitute PRM(µ) on E and (Yi) is an iid mark sequence with common distribution FY , independent of (Ti), the marked process of the points (Ti, Yi) is PRM(µ × FY).

Example. If Ti are the Poisson claim arrivals in a portfolio,

independent of the iid vectors Yi = (Xi, Di, Si) of claim sizes Xi,

delays in reporting Di and settlement times Si, then (Ti, Yi)

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Transformations of the points of a PRM yield a PRM

• If Ti are the points of a PRM(µ) on E then f (Ti) are the points of a PRM(µ) on f (E) with mean measure

µf(A) = µ({x ∈ E : f (x) ∈ A}) .

• One needs the additional restrictions that for bounded sets B ⊂ f (E), f−1(B) is bounded, or that µ is finite.

Example. Consider the marked PRM(µ × FX,D,S) on (0, ∞)4 of the points (Ti, Xi, Di, Si). The function

f (t, x, d, s) = (t, t + d, t + d + s, x)

is measurable and defines a PRM on (0, ∞)4 with points f (Ti, Xi, Di, Si) = (Ti, Ti + Di, Ti + Di + Si, Xi)

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The IBNR process as a model for teletraffic

• Consider a homogeneous Poisson arrival process on R

· · · T−2 < T−1 < 0 < T1 < T2 < · · · .

with intensity λ and an iid delay in reporting sequence (Di), independent of (Ti).

• The number

MT = NIBNR at time T = #{i ∈ Z0 : Ti ≤ T < Ti + Di}

= NT,D({(t, d) : t ≤ T < t + d}) .

counts the (non-observable at time T ) incurred but not reported (IBNR) claims.

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• For fixed T , MT is Poisson distributed, but the IBNR counting process (MT)T ≥0 is not a Poisson process.

• (MT)T ≥0 is a strictly stationary process with covariance

function

γ(h) = cov(NIBNR at time T , NIBNR at time T + h)

= λ

Z ∞

h

P (D > x) dx , h ≥ 0 .

Example. Assume Di is Pareto distributed:

P (D > x) = x−α , x ≥ 1 , α > 1 .

Then

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This function is not integrable for α ∈ (1, 2):

Z

1

γ(h) dh = ∞ .

Then the process M is said to have long memory with Hurst coefficient H = (3 − α)/2 ∈ (0.5, 1).

• The process M is a model for the activity in large computer networks (Internet), called the infinite source Poisson model.

• Ti is the arrival of an activity (packet) to the network and Di describes the amount of work brought into the system.

• The main object of interest is the workload

WT =

Z T

0

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• It describes the global performance of the network. The

behavior of the workload in the long memory case differs very much from the short memory case (lighter tail of D).

• Heavy tail components cause long memory in a stochastic system.

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PhD Course

Long memory, heavy tails and fractal behavior

Gennady Samorodnitsky (Cornell)

Copenhagen, Oct. 20-24

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Poisson integrals

• Let N = Pi επi be a PRM(µ) on E . Then the integral

Z E f dN = X i f (x)  X i επi(dx) = X i f (πi)

is defined as Lebesgue-Stieltjes integral (under conditions on f ).

• The integrals R fi dN , i = 1, 2, . . . are independent if the fi’s have disjoint support. (Inheritance from PRM)

• If fi ≥ 0, the integrals

R

fi dN , i = 1, 2, . . ., are independent if they are uncorrelated.

• Poisson integrals on a set of finite µ-measure are compound Poisson.

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Examples. The total claim amount

Z A×B x dNT,X(t, x) = X i:Ti∈A ,Xi∈B Xi .

The annual total claim amounts above and below a threshold u are independent: X i:Ti∈[0,1] XiI{Xi>u} and X i:Ti∈[0,1] XiI{Xi≤u} .

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Examples. The total claim amount Z A×B x dNT,X(t, x) = X i:Ti∈A ,Xi∈B Xi .

The annual total claim amounts above and below a threshold u are independent: X i:Ti∈[0,1] XiI{Xi>u} and X i:Ti∈[0,1] XiI{Xi≤u} . If N is PRM(Leb × µ) on [0, ∞) × Rd\{0} with R Rd\{0} min(1, |x|2) µ(dx) < ∞, then Yt = lim δ↓0 Z [0,t]×{x:δ<|x|≤1} x N (dt, dx) − Z [0,t]×{x:δ<|x|≤1} x dt µ(dx)  + Z [0,t]×{x:|x|>1} x N (dt, dx) , t ≥ 0 .

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Settled and outstanding RBNS claims.

• The ith claim is settled in [Ti + Di, Ti + Di + Si] according to a

deterministic non-decreasing continuous payment function f with f (0) = 0 and f (1) = 1.

• The settled reported but not settled (RBNS) claim amount at time T is given by Z {(t,x,d,s):t+d<T ≤t+d+s} x f (T − t − d)/s = X i::Ti+Di<T ≤Ti+Di+Si Xi f (T − Ti − Di)/Si d = M X i=1 Xi f Πi  .

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• The distributions of the settled and outstanding RBNS claim

amounts are compound Poisson and can be simulated or

calculated numerically.

• In this way, one can also approximate the distribution of reserves (predict).

The chain ladder.

• Assume

Ni,i+j , j = 0, 1, . . . , m − i

are the counts of payments for claims that occurred in the ith year and for which payments were executed in the years

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• A simplistic idea of predicting Ni,m+1 is to assume that the best predictor (in the mean square sense) has form

(Mack) E(Ni,m+1 | Nii, . . . , Nim) = fm−i Nim

for some constants fj ≥ 1 which are estimated by the chain ladder estimators bfj.

• The quantity Nci,m+1 = bfm−i Nim is then not the best predictor

of Ni,m+1 and determining the prediction error is a problem.

• Such a model is hard to simulate because of the lack of

assumptions. Although counts of points underly the model it is difficult (impossible?) to find a point process model for Ni,i+j

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Can one predict (in the mean square sense) in a point process model?

• Assume that claims arrive at the times 0 < T1 < T2 < · · · of a homogeneous Poisson process on (0, ∞) with intensity λ.

• The ith claim triggers a stream of payments Xik at the times

Ti + Yik, k = 1, 2, . . ..

• For every i, (Yik)k≥1 constitutes a homogeneous Poisson process on (0, ∞) with intensity γ, independent of (Xik)k≥1.

• The payments Xik are iid.

• The points ((Yik)k≥1, (Xik)k≥1) constitute an iid sequence,

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• This means that the points (Ti, (Yik)k≥1, (Xik)k≥1) constitute a marked PRM N on a suitable state space E.

• The Poisson integrals

M (0, b] = Z E ∞ X k=1 I{t∈[0,1], t+yk∈[0,b]} N (dt, d(yk), d(xk)) = NT[0,1] X i=1 ∞ X k=1 I{Ti+Yik∈[0,b]} S(0, b] = Z E ∞ X k=1 xk I{t∈[0,1], t+yk∈[0,b]} N (dt, d(yk), d(xk)) = NT[0,1] X i=1 ∞ X k=1 Xik I{Ti+Yik∈[0,b]}

describe the number of payments/paid amounts for claims

arriving in the first year [0, 1] and payments being executed in [0, b] for some b ≥ 1.

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• For b ≥ 1, x > 0, we can calculate the mean square predictors of M (b, b + x] and S(b, b + x] given M (0, b]: E(M (b, b + x] | M (0, b] = m) = λ γ x φ (m) 1 (γ) φ(m)2 (γ) E(S(b, b + x] | M (0, b] = m) = EX11 λ γ x φ (m) 1 (γ) φ(m)2 (γ) ,

where φ(m)1 , φ(m)2 are the mth derivatives of the Laplace-Stieltjes transforms of the random variables PL+1i=1 (b − Ui), PLi=1(b − Ui),

respectively, and L is a Poisson(λ) distributed random variable independent of the iid uniform U(0, 1) sequence (Ui).

• This model does not satisfy the condition (Mack) of the chain ladder model.

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References

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